Tag Archives: Kinematics in Two Dimensions

College Physics by Openstax Chapter 2 Problem 62

By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s2 at t=10 s .

Figure 2.63

Solution:

To solve for the slope of the curve at t=10 s, we need two points — 1 just before and 1 just after.

When t=0 \ \text{s}, v=166 \ \text{m/s} and when t=20 \ \text{s}, v=230 \ \text{m/s}. Therefore, the acceleration is

\begin{align*}
\text{acceleration} & =\text{slope} \\
\text{acceleration} & =\frac{\Delta v}{\Delta t} \\
\text{a} & = \frac{v_2-v_1}{t_2-t_1} \\
\text{a} & = \frac{230\ \text{m/s}-166\ \text{m/s}}{20\ \text{s}-0\ \text{s}} \\
\text{a} & =3.2\ \text{m/s}^2 \\

\end{align*}

Note that the values are approximated to satisfy the given acceleration in the problem statement. The values may differ from one’s answer due to some uncertainties of a graph.

Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 58

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10−3s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Solution:

The concept is the same with Problem 2.56.

Part A

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

 v_{y_2} & = \sqrt{\left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right)} \\

 v_{y_2} & = -  \sqrt{\left( 0 \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m} - 1.50\ \text{m} \right)} \\

 v_{y_2} & = - 5.42 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}



\end{align*}

Part B

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

\left( v_{y_1} \right)^2 & = \left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right) \\

v_{y_1} & = \sqrt{\left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right)} \\

v_{y_1} & = \sqrt{\left( 0 \ \text{m/s} \right)^2 -2\left( -9.81 \ \text{m/s}^2  \right) \left( 1.10 \ \text{m} - 0 \ \text{m}  \right) }\\

v_{y_1} & =  4.65 \ \text{m/s}  \qquad {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

Part C

\begin{align*}

a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_2-v_1}{\Delta t} \\
a & = \frac{4.65 \ \text{m/s} - \left( -5.42 \ \text{m/s} \right)}{3.50 \times 10^{-3} \ \text{s}} \\
a & = 2877 \ \text{m/s}^2 \\
a & = 2.88 \times 10^{3}\    \text{m/s}^2 \ \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part D

\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\Delta y & =  \frac{\left( v_{y_2} \right)^2 -  \left( v_{y_1} \right)^2}{2 a} \\
\Delta y & = \frac{\left( 0 \ \text{m/s} \right)^2 - \left( -5.42 \ \text{m/s} \right)^2}{2\left( 2.88 \times 10^3 \ \text{m/s}^2 \right)}\\
\Delta y & = -0.00510 \ \text{m} \\
 \Delta y & = -5.10 \times 10^{-3} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 57

A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.

Solution:

Part A

Figure A

Consider Figure A.

We are interested in two positions. Position 1 is where the coin is dropped. At this position, the coin is 300 m above the ground, the time is 0 s, and the velocity is 10.0 m/s upward.

Position 2 is the highest point of the coin reaches. At this position, the velocity is equal to 0 m/s.

Position 1 is the initial position and position 2 is the final position. Solve for the value of y2.

\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 +2a \Delta y \\
\Delta y & = \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2 - y_1 & =  \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} \\
y_2& =  \frac{\left( v_{y_2} \right)^2 - \left( v_{y_1} \right)^2}{2a} +y_1 \\
y_2 & = \frac{\left( 0 \ \text{m/s} \right)^2-\left( 10.0\ \text{m/s} \right)^2}{2\left( -9.81 \ \text{m/s}^2 \right)}+300\ \text{m}
\\
y_2 & =305 \ \text{m} \ \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}

\therefore The maximum height reached by the coin is about 305 meters from the ground.

Part B

We do not know the position 4 seconds after the coin has been released, the answer can be above or below the initial point. We can actually use one of the kinematical equations to solve for the final position given the time. Here, the initial position is the point of release and the final position is the point of interest at 4.00 seconds after release.

\begin{align*}
\Delta y & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 - y_1 & = v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = y_1 +  v_{y_1}t+\frac{1}{2}at^2 \\
y_2 & = 300\ \text{m} +\left( 10\ \text{m/s} \right)\left( 4.00\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 4.00\ \text{s} \right)^2\\
y_2 & = 261.52\ \text{m} \\
y_2 & = 262\ \text{m}\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}  \\
\end{align*}

\therefore The coin is at a height of 262 meters above the ground 4.00 seconds after release. That is, the coin is already dropping and it is already below the release point.

Solving for the velocity 4.00 seconds after release considering the same initial and final position.

\begin{align*}
v_{y_2} & = v_{y_1}+at \\
v_{y_2} & = 10\ \text{m/s} + \left( -9.81\ \text{m/s}^2 \right)\left( 4.00 \ \text{s} \right) \\
v_{y_2} & = -29.2\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\therefore The coin has a velocity of 29.2 m/s directed downward 4.00 seconds after it is released. This confirms that the coin is indeed moving downwards at this point.

Part C

Figure C

Considering figure C, we have two positions. Position 1 is the point of release 300 m above the ground with a velocity of 10 m/s upward. This is time 0 s.

The second position is at the ground where y=0 m. We are interested at the time in this position.

Considering position 1 as the initial position and position 2 as the final position.

\begin{align*}
\Delta y & = v_{y_1} \Delta t+\frac{1}{2}a\left( \Delta t \right)^2 \\
y_2-y_1 & =  v_{y_1}t+\frac{1}{2}at^2 \\
0\ \text{m}-300\ \text{m} & = \left( 10 \ \text{m/s} \right)t+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)t^2 \\
-300 & = 10t-4.905t^2 \\
4.905t^2-10t-300 & = 0 \\
\end{align*}

Solve for the value of t using the quadratic formula with a=4.905, b=-10, and c=-300.

\begin{align*}
t & = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\
t & = \frac{-\left( -10 \right) \pm \sqrt{\left( 10 \right)^2-4\left( 4.905 \right)\left( -300 \right)}}{2\left( 4.905 \right)}\\
t & = 8.91 \ \text{s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\therefore The time is about 8.91 seconds before the coin hits the ground.

Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 56

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (8.00×10−5 s) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Solution:

Part A

Figure A

For this part, we shall consider Figure A.

We will be considering the two positions as shown. The first position is when the ball is dropped from a height of 1.50 meters. For this position, we know that y1=1.50 m, t1=0 s, and vy1=0 m/s.

Position 2 is immediately after the ball hits the floor. For this position, we do not the time elapse and the velocity but we know that the height is zero. That is y2=0 m.

Position 1 is the initial position and position 2 is the final position. Solving for vy2, we have

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

 v_{y_2} & = \sqrt{\left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right)} \\

 v_{y_2} & = -  \sqrt{\left( 0 \right)^2+2\left( -9.81\ \text{m/s}^2 \right)\left( 0\ \text{m} - 1.50\ \text{m} \right)} \\

 v_{y_2} & = - 5.42 \ \text{m/s} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}



\end{align*}

\therefore The steel ball has a velocity of about 5.42 m/s directed downward when it strikes the floor.

Part B

Figure B

Figure B shows the two positions we are interested in to solve for this part.

Position 1 is at the floor immediately just after the ball hits it. At this initial position, we have y1=0 m, and t1= 0 s. We do not know the velocity at this point.

At position 2, the ball bounced back to its second peak. We know that at the peak of a free falling body, the velocity is zero. So, for this final position, we have y2=1.45 m, and vy2=0 m/s. We do not know the time at this position.

Position 1 is the initial position while position 2 is the final position. Solving for the initial velocity we have

\begin{align*}

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\

\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \left( y_2 - y_1  \right) \\

\left( v_{y_1} \right)^2 & = \left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right) \\

v_{y_1} & = \sqrt{\left( v_{y_2} \right)^2 - 2 a \left( y_2 - y_1  \right)} \\

v_{y_1} & = \sqrt{\left( 0 \ \text{m/s} \right)^2 -2\left( -9.81 \ \text{m/s}^2  \right) \left( 1.45 \ \text{m} - 0 \ \text{m}  \right) }\\

v_{y_1} & =  5.33 \ \text{m/s}  \qquad {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

\therefore The steel ball has a velocity of about 5.33 m/s directed upward immediately after it leaves the floor.

Part C

From our answers in Part A and Part B, we have a change in velocity from -5.42 m/s to 5.33 m/s. So, in this case the initial velocity is v1=-5.42 m/s and the final velocity is v2=5.33 m/s. We can compute for the acceleration:

\begin{align*}

a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_2-v_1}{\Delta t} \\
a & = \frac{5.33 \ \text{m/s} - \left( -5.42 \ \text{m/s} \right)}{8.00 \times 10^{-5} \ \text{s}} \\
a & = 134,375 \ \text{m/s}^2 \\
a & = 1.34 \times 10^{5}   \text{m/s}^2 \ \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part D

The period of compression happens when the ball has a velocity of -5.42 m/s until it reaches 0 m/s. We shall solve for the change in displacement for this two given velocities. The initial velocity is -5.42 m/s and the final velocity is 0 m/s. The acceleration during this period is the one solved in Part C, a=1.34×105 m/s2. 

\begin{align*}
\left( v_{y_2} \right)^2 & = \left( v_{y_1} \right)^2 + 2 a \Delta y \\
\Delta y & =  \frac{\left( v_{y_2} \right)^2 -  \left( v_{y_1} \right)^2}{2 a} \\
\Delta y & = \frac{\left( 0 \ \text{m/s} \right)^2 - \left( -5.42 \ \text{m/s} \right)^2}{2\left( 1.34 \times 10^5 \ \text{m/s}^2 \right)}\\
\Delta y & = -0.000110 \ \text{m} \\
 \Delta y & = -1.10 \times 10^{-4} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\
\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 55

Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return.
(a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s.
(b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.

Solution:

Part A

Figure A

Consider Figure A.

We shall consider two points for our solution. First, position 1 is the top of the well. In this position, we know that y1=0, t1=0 and vy1=0.

Position 2 is located at the top of the water table where the rock will meet the water. Since we neglect the time for the sound to travel from position 2 to position 1, we can say that t2=2.0000 s, the time of the rock to reach this position.

Solving for the value of y2 will determine the distance between the two positions.

\begin{align*}

\Delta y & = v_{y_1}t+\frac{1}{2}at^2  \\
y_2-y_1 & = v_{y_1}t+\frac{1}{2}at^2  \\
y_2 & = y_1 +v_{y_1}t+\frac{1}{2}at^2  \\
y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 2.0000\ \text{s} \right)^2 \\
y_2 & = -19.6\ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer }\right)}

\end{align*}

Position 2 is 19.6 meters measured downward from position 1.

\therefore The distance to the water is about 19.6 meters.

Part B

For this case, the 2.0000 seconds that is given includes the time that the rock travels from position 1 to position 2, tr, and the time that the sound travels from position 2 to position 1, ts.

\begin{align*}
t_r+t_s & =2.0000\ \text{s} \\
t_s & = 2.0000\ \text{s}-t_r 
\end{align*}

Considering the motion of the rock from position 1 to position 2.

\begin{align*}

\Delta y & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2  \\
y_2-y_1 & = v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2  \\
y_2 & = y_1 +v_{y_1}t_r+\frac{1}{2}a\left( t_r \right)^2  \\
y_2 & = 0+0+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( t_r \right)^2  \\
y_2 & = -4.905\left( t_r \right)^2  \qquad {\color{Blue} \text{Equation 1}}\\


\end{align*}

Now, let us consider the motion of the sound from position 2 to position 1. Sound is assumed to have a constant velocity of 322.00 m/s.

\begin{align*}

\Delta y & = v_s \times t_s \\
y_1-y_2 & =\left( 322.00\ \text{m/s} \right)\left( t_s \right) \\
0-y_2 & =\left( 322.00 \right)\left( 2.0000-t_r \right) \\
y_2 & = -322.00\left( 2.0000-t_r \right) \qquad  {\color{Blue} \text{Equation 2}}


\end{align*}

So, we have two equations from the two motions. We can solve the equations simultaneously. 

\begin{align*}

-4.905 \left( t_r \right)^2 & = -322.00\left( 2.0000-t_r \right) \\
4.905 \left( t_r \right)^2 & =322.00\left( 2.0000-t_r \right) \\
4.905 \left( t_r \right)^2 & = 644.00-322.00t_r \\
4.905\left( t_r \right)^2 + 322.00t_r-644.00 & = 0 \\

\end{align*}

We can solve the quadratic formula using the quadratic equation. 

\begin{align*}

t_r & = \frac{-b \pm\sqrt{b^2-4ac}}{2a} \\
t_r & = \frac{-322.00\pm\sqrt{\left( 322.00 \right)^2-4\left( 4.905 \right)\left( -644.00 \right)}}{2\left( 4.905 \right)}\\
t_r & =1.9425 \ \text{s}

\end{align*}

Now that we have solved for the value of tr, we can use this to solve for y2 using either Equation 1 or Equation 2. We will use equation 1.

\begin{align*}

y_2 & = -4.905\left( t_r \right)^2 \\
y_2 & = -4.905 \left( 1.9425 \right)^2 \\
y_2 & =-18.5 \ \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

Position 2 is about 18.5 meters below position 1.

\therefore In this case, the distance between the two positions is 18.5 meters.

Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 54

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past the window. What was the ball’s initial velocity? Hint: First consider only the distance along the window, and solve for the ball’s velocity at the bottom of the window. Next, consider only the distance from the ground to the bottom of the window, and solve for the initial velocity using the velocity at the bottom of the window as the final velocity.

Solution:

First, we have the position 1 where the motion starts. Here, we know that y<sub>1</sub>=0, t<sub>1</sub>=0, and v<sub>y1</sub>=0. Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y<sub>2</sub>=7.50 meters. We do not know the time and velocity at this point. Then we have position 3 at the top of the window where the overall height is 9.50 meters, y<sub>3</sub>=9.50. We also do not know the velocity and time elapsed in this position.
Figure A

Consider Figure A. We shall be considering the three positions shown.

First, we have position 1 where the motion starts. Here, we know that y1=0 and t1=0, but we do not know vy1.

Position 2 is at the bottom of the window. We know that it is 7.50 meters from where the motion started. So we have y2=7.50 meters. We do not know the time and velocity at this point.

Then we have position 3 at the top of the window where the overall height is 9.50 meters, y3=9.50. We also do not know the velocity and time elapsed in this position.

Consider positions 2 and 3. The initial position in this case is at position 2 and the final position is at position 3. We know that the difference of time between this two positions is 0.312 seconds. We can say that

t_3 =t_2+0.312 \ \text{s} \\
t_3-t_2 = 0.312\ \text{s}

Using the same 2 positions still, we have

\begin{align*}

y_3 & = y_2 + v_{y_2} \Delta t+\frac{1}{2}a\left(  \Delta t \right)^2 \\
9.50\ \text{m} & = 7.50\ \text{m} +  v_{y_2} \left( t_3-t_2 \right)+\frac{1}{2}a\left( t_3-t_2 \right)^2 \\
9.50\ \text{m}-7.50\ \text{m} & = v_{y_2}\left( 0.312\ \text{s} \right)+\frac{1}{2}\left( -9.81\ \text{m/s}^2 \right)\left( 0.312\ \text{s} \right)^2\\
2.00\ \text{m} & = 0.312\ \text{s} \left( v_{y_2} \right)-0.4775\ \text{m} \\
 0.312\ \text{s} \left( v_{y_2} \right) & = 2.00\ \text{m}+0.4775\ \text{m} \\
 0.312\ \text{s} \left( v_{y_2} \right) & = 2.4775\ \text{m} \\
v_{y_2}& =\frac{2.4775\ \text{m}}{0.312\ \text{s}} \\
v_{y_2}& = 7.94\ \text{m/s}

\end{align*}

We have computed the velocity of the ball at the bottom of the window.

Next, we shall consider positions 1 and 2. In this consideration, position 1 will be considered the initial position while position2 is the final position. 

\begin{align*}

\left( v_{y_2} \right)^2  &  = \left( v_{y_1} \right)^2 +2a \Delta y \\
\left( 7.94\ \text{m/s} \right)^2 & = \left( v_{y_1} \right)^2 + 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\
\left( v_{y_1} \right)^2 & = \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)\\
v_{y_1} & = + \sqrt{ \left( 7.94\ \text{m/s} \right)^2- 2\left( -9.81 \ \text{m/s}^2 \right)\left( 7.50\ \text{m}-0 \right)} \\
v_{y_1} & = + 14.5\ \text{m/s}\qquad {\color{DarkOrange} \left( \text{Answer} \right) }

\end{align*}

\therefore The ball’s initial velocity is about 14.5 m/s upward.

Advertisements
Advertisements

College Physics by Openstax Chapter 3 Problem 36

The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

Solution:

We are required to solve for the maximum distance. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

  • The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
  • The jumper is on level ground, and the motion started from the ground.

The formula for range is

\text{R}=\frac{\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}}{\text{g}}

Since we are already given the necessary details, we can now solve for the range.

\begin{align*}
 \text{R}&=\frac{\left(9.5\:\text{m/s}\right)^2\:\sin 90^{\circ} }{9.81\:\text{m/s}^2}\\
\text{R}&=9.20\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ 
\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 3 Problem 33

The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37×103 km . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

Solution:

Part A

We are given the range of the projectile motion. The range is 32.0 km. We also know that for the projectile to reach its maximum distance, it should be fired at 45°. So from the formula of range,

\displaystyle \text{R}=\frac{\text{v}_0^2\:\sin 2\theta _0}{\text{g}}

we can say that \sin 2\theta _0=\sin \left(2\times 45^{\circ} \right)=\sin 90^{\circ} =1. So, we have

\displaystyle \text{R}=\frac{\text{v}_0^2}{\text{g}}

We can solve for v0 in terms of the other variables. That is

\displaystyle \text{v}_0=\sqrt{\text{gR}}

Substituting the given values, we have

\begin{align*}
\displaystyle \text{v}_0 & =\sqrt{\left(9.81\:\text{m/s}^2\right)\left(32\times 10^3\:\text{m}\right)} \\
\displaystyle \text{v}_0 & =560.29\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Part B

We are solving for the maximum height here, which happened at the mid-flight of the projectile. The vertical velocity at this point is zero. Considering all this, the formula for the maximum height is derived to be

\displaystyle \text{h}_{\text{max}}=\frac{\text{v}_{0_y}^2}{2\text{g}}

The initial vertical velocity, v0y, is calculated as

\begin{align*}
\text{v}_{\text{0y}} & =\text{v}_0\sin \theta _0 \\
& =\left(560.29\:\text{m/s}\right)\sin 45^{\circ}  \\
& =396.18\:\text{m/s}
\end{align*}

Therefore, the maximum height is

\begin{align*}
\text{h}_{\text{max}} & =\frac{\left(396.18\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)} \\
\text{h}_{\text{max}} & =8000\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\

\end{align*}

Part C

Consider the following figure

A right triangle is formed with the legs, the horizontal distance and the radius of the earth, and the hypotenuse is the sum of the radius of the earth and the distance d, which is the unknown in this problem. Using Pythagorean Theorem, and converting all units to meters, we have

\begin{align*}
\text{R}^2+\left(32.0\times 10^3\:\text{m}\right)^2 & =\left(\text{R}+\text{d}\right)^2 \\
\left(6.37\times 10^6\:\text{m}\right)^2+\left(32.0\times 10^3\:\text{m}\right)^2 & =\left(6.37\times 10^6+\text{d}\right)^2 \\
\text{d} & =\sqrt{\left(6.37\times \:10^6\:\right)^2+\left(32.0\times 10^3\:\right)^2}-6.37\times \:10^6 \\
\text{d} & =80.37\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

This error is not significant because it is only about 1% of the maximum height computed in Part B.

Advertisements
Advertisements

College Physics by Openstax Chapter 3 Problem 32

Verify the ranges shown for the projectiles in Figure 3.40(b) for an initial velocity of 50 m/s at the given initial angles.

Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial angles. To do this, we shall use the formula

\displaystyle \text{R}=\frac{\text{v}_{\text{0}}^2 \sin 2\theta _{\text{0}}}{\text{g}}

When the initial angle is 15°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 15^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

When the initial angle is 45°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}

When the initial angle is 75°, the range is

\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 75^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

Based on the result of the calculations, we can say that the numbers in the figure are verified. The very small differences are only due to round-off errors.

Advertisements
Advertisements

College Physics by Openstax Chapter 3 Problem 11

Find the components of vtot along the x- and y-axes in Figure 3.55.

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis.
Figure 3.55

Solution:

By isolating the vtot from the rest of the other vectors, we come up with the following figure. Also, the x and y-components are shown.

The resultant velocity and its x and y components

The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. To solve for the x and y components, we just need to solve the legs of the right triangle formed by the three vectors. That is, 

 \text{x-component}=\left(6.72\:\text{m/s}\right)\cos 49^{\circ} =4.41\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\text{y-component}=\left(6.72\:\text{m/s}\right)\sin 49^{\circ} =5.07\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements