Tag Archives: Openstax Solutions

College Physics by Openstax Chapter 6 Problem 28

Riding a Bicycle in an Ideally Banked Curve

Problem:

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.33. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that \theta (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is, \theta = \tan ^{-1} \left( v^2/rg \right)

(b) Calculate \theta for a 12.0 m/s turn of radius 30.0 m (as in a race).

Figure 6.33 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system, while its horizontal component must supply the centripetal force. This process produces a relationship among the angle θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.

Solution:

Part A

Let us redraw the given forces in a free-body diagram with their corresponding components.

The force N and F_c are the vertical and horizontal components of the force F.

If we take the equilibrium of forces in the vertical direction (since there is no motion in the vertical direction) and solve for F, we have

\begin{align*}
\sum F_y & = 0 \\ \\
F \cos \theta - mg & = 0 \\ \\
F \cos \theta & = mg \\ \\
F & = \frac{mg}{\cos \theta}  \quad \quad  & \color{Blue}  \small \text{Equation 1}
\end{align*}

If we take the sum of forces in the horizontal direction and equate it to mass times the centripetal acceleration (since the centripetal acceleration is directed in this direction), we have

\begin{align*}
\sum F_x & = ma_c \\ \\
F \sin \theta  & = m a_c \\ \\
F \sin \theta  & = m \frac{v^2}{r}   \quad \quad  & \color{Blue}  \small \text{Equation 2}
\end{align*}

We substitute Equation 1 to Equation 2.

\begin{align*}
F \sin \theta  & = m \frac{v^2}{r} \\ \\
\frac{mg}{\cos \theta} \sin \theta & = m \frac{v^2}{r} \\ \\
mg \frac{\sin \theta}{\cos \theta} & =m \frac{v^2}{r} \\ \\
\end{align*}

We can cancel m from both sides, and we can apply the trigonometric identity \displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We should come up with

\begin{align*}
g \tan \theta & = \frac{v^2}{r} \\ \\
\tan \theta & = \frac{v^2}{rg} \\ \\
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are given the following values:

  • linear velocity, v = 12.0\ \text{m/s}
  • radius of curvature, r=30.0\ \text{m}
  • acceleration due to gravity, g = 9.81\ \text{m/s}^2

We substitute the given values to the formula of \theta we solve in Part A.

\begin{align*}
\theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \\ \\
\theta & = \tan ^ {-1} \left[ \frac{\left( 12.0\ \text{m/s} \right)^2}{\left( 30.0\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\
\theta & = 26.0723 ^\circ \\ \\
\theta & = 26.1 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 6 Problem 24

Centripetal Force of a Rotating Wind Turbine Blade

Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

Solution:

We are given the following values:

  • radius, r=100\ \text{m}
  • angular velocity, \omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}
  • mass, m=4\ \text{kg}

Centripetal force F_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v and has magnitude F_c = m a_c which can also be expressed as

F_c = m \frac{v^2}{r} \quad \text{or} \quad \ F_c = mr \omega^2

For this particular problem, we are going to use the formula F_c = mr \omega^2. If we substitute the given values, we have

\begin{align*}
F_c & =mr \omega^2 \\ \\
F_c & = \left( 4\ \text{kg} \right)\left( 100\ \text{m} \right)\left( 3.1416\ \text{rad/sec} \right)^2 \\ \\
F_c & = 3947.8602\ \text{N} \\ \\
F_c & = 4 \times 10^3\ \text{N}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The centripetal force on the end of the wind turbine blade is approximately 4 \times 10^3\ \text{N}.

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Problem 6-19: The angular velocity of an “artificial gravity”

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?

Solution:

We are given the following quantities:

\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

\begin{align*}
\omega & = \sqrt{\frac{a_c}{r}}  \\ \\
\omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\
\omega & = 0.313 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Problem 6-18: The linear speed of an ultracentrifuge and Earth in its orbit

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Solution:

Part A

We are given a linear speed of an ultracentrifuge of 0.50\ \text{km/s}. We are asked to verify this value if we are given a radius of r=0.100\ \text{m} and angular velocity of \omega = 50000 \ \text{rev/min}. We are going to use the formula

v = r \omega

Since we are given a linear speed in \text{km/s}, we are going to convert the radius to \text{km}, and the angular velocity to \text{rad/sec}

r=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km} 
\omega = 50000 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} =5235.9878\ \text{rad/sec}

Now, we can substitute these into the formula

\begin{align*}
v & = r \omega \\ \\
v & = \left( 0.0001 \ \text{km} \right)\left( 5235.9878 \ \text{rad/sec} \right) \\ \\
v & = 0.5236 \ \text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This value is about 0.500 km/s.

Part B

From Table 6.2 of the book

ParentSatelliteAverage orbital radius r(km)Period T(y)r3 / T2 (km3 / y2)
SunEarth1.496 \times 10^{8} 13.35 \times 10^{24}

Using the same formulas we used in Part A, we can solve for the linear velocity of the Earth around the sun. The radius is

r=1.496 \times10^{8} \ \text{km}

The angular velocity is

\begin{align*}
\omega &  = 1 \ \frac{\text{rev}}{\text{year}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{year}}{365.25 \ \text{days}} \times \frac{1\ \text{day}}{24\ \text{hours}}\times \frac{1\ \text{hour}}{3600\ \text{sec}} \\ \\ 
\omega &  = 1.9910 \times 10^{-7}\ \text{rad/sec}
\end{align*}

The linear velocity is

\begin{align*}
v & = r \omega \\ \\
v & = \left( 1.496\times 10^{8}\ \text{km} \right)\left( 1.9910 \times 10 ^ {-7} \right) \ \text{rad/sec}\\ \\
v & = 29.7854\ \text{km/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The linear velocity is about 30 km/s.

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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.

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College Physics by Openstax Chapter 2 Problem 61

Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s.

Figure 2.62

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertainties when using graphs. The difference is not really significant for this case.

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College Physics by Openstax Chapter 2 Problem 60

Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s. Assume all values are known to 2 significant figures.

Figure 2.62

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertain when using graphs. The difference is not really significant for this case.

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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.

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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.

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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.

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