Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

Assuming that the gun was fired horizontally.

Consider the horizontal component of the motion.

v_{o_x}=1100\:m/s

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

Consider the vertical component.

v_{o_y}=0\:m/s

\Delta y=V_{0_y}t-\frac{1}{2}gt^2=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(0.32\:s\right)^2=-0.50

The negative sign of  \Delta y means that the bullet went downward.

Therefore, the bullet traveled 352 m horizontally and 0.50 m vertically downward in 0.32 seconds.

Grantham PHY220 Week 2 Problem 2

An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y axis and the total distance to the plane from you. Choose the x-axis east, y axis north, and z axis up.

Solution:

The angle of the plane in the x-y axis

Solve for \theta

\theta =tan^{-1}\left(\frac{1500}{2100}\right)=35.5^{\circ}

The angle is 35.5 degrees measured from the west axis. If it is measured from the East axis, the angle would be 180-35.5=144.5 degrees.

The distance from the plane to the observer

Solve for d=\sqrt{\left(1500\:m\right)^2+\left(2100\:m\right)^2+\left(5000\:m\right)^2}=5626.72\:m

College Physics 2.49 – A ball passing a tree branch


You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?


Solution:

The known values are

a=-9.80\:m/s^2;\:v_o=15.0\:m/s;\:y=7.00\:m

The applicable formula is y=v_ot+\frac{1}{2}at^2.

Using this formula, we can solve it in terms of time, t.

t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}

Substituting the known values, we have

t=\frac{-15.0\:m/s\pm \sqrt{\left(15.0\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(7.00\:m\right)}}{-9.80\:m/s^2}

t=\frac{-15.0\:m/s\pm 9.37\:m/s}{-9.80\:m/s^2}

We have two values for time, t. These two values represent the times when the ball passes the tree branch. So, the two values of the time, t, are  

t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec

t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds. That is 1.92 seconds.


College Physics 2.48 – A very strong, but inept, shot putter


A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?


Solution:

The known values are

\text{y}_0=2.20\:\text{m};\:\text{y}=1.80\:\text{m};\:\text{v}_0=11.0\:\text{m/s};\:\text{and}\:\text{a}=-9.80\:\text{m/s}^2

The applicable formula is \Delta y=v_0t+\frac{1}{2}at^2.

1.80\:m-2.20\:m=\left(11.0\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2

-0.40\:m=\left(11.0\:m/s\right)t-\left(4.90\:m/s^2\right)t^2

4.90t^2-11t-0.40=0

Using the quadratic formula solve for t, we have

\:t=\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)}

t=2.28\:sec\:\:\:\:\:or\:\:\:\:\:\:t=-0.04

Therefore, t=2.28\:s


College Physics 2.47 – Rock thrown from a cliff


(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.

(b) How long would it take to reach the ground if it is thrown straight down with the same speed?


Solution:

Part A

Refer to the figure below.

The known values are: \text{t}=2.35\:\text{s};\:\text{y}=0\:\text{m};\:\text{v}_0=+8.00\:\text{m/s};\:\text{a}=-9.8\:\text{m/s}^2

Based from the given values, we can use the equation y=y_0+v_0t+\frac{1}{2}at^2. Substituting the values, we have

0\:=\text{y}_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2

y=8.26\:m

Therefore, the cliff is 8.26 meters high.

Part B

Refer to the figure below

The knowns now are: y=0\:m;\:y_0=8.26\:m;\:v_0=-8.00\:m/s;\:a=-9.80\:m/s^2

Based from the given values, we can use the equation y=y_0+v_0t+\frac{1}{2}at^2. Substituting the values, we have

0\:m=8.26\:m+\left(-8.00\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2

4.9\text{t}^2+8\text{t}-8.26=0

Using the quadratic formula to solve for the value of t, we have

\displaystyle \text{t}=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)}

\text{t}=0.717\:\text{s}


College Physics 2.46 – A swimmer bouncing from a diving board


A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?


Solution:

The known values are: y_0=1.80\:m,\:y=0\:m,\:a=-9.80\:m/s^2,\:v_0=4.00\:m/s

Part A

Based from the knowns, the formula most applicable to solve for the time is  \Delta y=v_0t+\frac{1}{2}at^2. If we rearrange the formula by solving for t, and substitute the given values, we have

t=\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a}

t=\frac{-4.00\:m/s\pm \sqrt{\left(4.00\:m/s\right)^2-2\left(-9.80\:m/s^2\right)\left(1.80\:m\right)}}{-9.80\:m/s^2}

t=1.14\:s

Part B

We have the formula

\Delta y=\frac{v^2-v_0^2}{2a}

\Delta y=\frac{\left(0\:m/s\right)^2-\left(4.00\:m/s\right)^2}{2\left(-9.80\:m/s^2\right)}

\Delta y=0.816\:m

Part C

The formula to be used is

v^2=v_0^2+2a\Delta y

v=\pm \sqrt{v_0^2+2a\Delta y}

v=\pm \sqrt{\left(4.00\:m/s\right)^2+2\left(-9.80\:m/s^2\right)\left(-1.80\:m\right)}

v=\pm \sqrt{51.28\:m^2/s^2}

v=\pm 7.16\:m/s

Since the diver must be moving in the negative direction, v=-7.16\:m/s


College Physics 2.45 – A dolphin in an aquatic show


A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

(a) List the knowns in this problem.

(b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable.

(c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.


Solution:

Part A

The knowns are 

a=-9.80\:m/s^2

v_0=13\:m/s

y=0\:m

Part B

At the highest point of the jump, the velocity is equal to 0. That is v=0\:m/s

Basing from the known values, the formula that we can use to solve for y is v^2=v_0^2+2a\Delta y. By rearranging these variables, the formula in solving for \Delta y is \Delta y=\frac{v^2-v_0^2}{2a}. Therefore we have, 

\Delta y=\frac{\left(0\:m/s\right)^2-\left(13.0\:m/s\right)^2}{2\left(-9.80\:m/s^2\right)}

\Delta y=8.62\:m

This value is reasonable since dolphins can jump several meters high out of the water. Usually, a dolphin measures about 2 meters and they can jump several times their length.

Part C

The formula we can use to solve for the time is v=v_0+at. If we rearrange this formula and solve for t, it becomes  t=\frac{v-v_0}{a}.

t=\frac{0\:m/s-13.0\:m/s}{-9.8\:m/s^2}

t=1.3625\:s

This value is the time it takes the dolphin to reach the highest point. Since the time it takes to reach this point is equal to the time it takes to go back to water, the time it is in the air is 2.65\:s.


College Physics 2.44 – A rescue helicopter over a sunk boat


A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.


Solution:

Part A

The knowns are 

a=-9.80\:m/s^2

v_0=-1.40\:m/s

t=1.8\:s

y=0\:m

Part B

y_0=-y+v_0t+\frac{1}{2}at^2

y_0=0\:m\:+\left(-1.40\:m/s\right)\left(1.8\:s\right)+\frac{1}{2}\left(-9.80\:m/s^2\right)\left(1.8\:s\right)^2

y_0=18\:m