Freely Falling Objects Problems and Solutions

Q#37  

(I) A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.55 s. How high is the cliff? 

Answer:

Choose downward to be the positive direction, and take $y_0$ = 0 at the top of the cliff. 

The initial velocity is $v_0$ = 0, and the acceleration is a = $9.80 \ m/s^2$. 

The displacement is found from Eq. $y=y_0+v_{0y}t+\frac{1}{2}at^2$ with x replaced by y. 

$y=0+0+\frac{1}{2}(9.8 \ m/s^2)(3.55 \ s)^2$

$y=61.8 \ m$

Q#38

(I) Estimate 

(a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and 

(b) his velocity just before “landing.”

Answer:

Choose downward to be the positive direction, and take $y_0$ = 0 to be at the top of the Empire State Building. The initial velocity is $v_0$ = 0, and the acceleration is a = -9.80 $m/s^2$.

(a) The elapsed time can be found from Eq. $y=y_0+v_{0y}t+\frac{1}{2}at^2$, 

with x replaced by y. 

$y=0+0+\frac{1}{2}at^2$

$t=\sqrt{\frac{2y}{a}}$

$t=\sqrt{\frac{2(380 \ m)}{9.8 \ m/s^2}}=8.806 \ s$

(b) The final velocity can be found from Eq. 

$v_y=v_{0y}+at$

$v_y=0+(9.80 \ m/s^2)(8.806 \ s)=86 \ m/s$

Q#39

(II) A ball player catches a ball 3.4 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach? 

Answer:

Choose upward to be the positive direction, and take $y_0$ = 0 to be the height from which the ball was thrown. The acceleration is a = −9.80 $m/s^2$ . 

The displacement upon catching the ball is 0, assuming it was caught at the same height from which it was thrown. The starting speed can be found from Eq. 

$y=y_0+v_{0y}t+\frac{1}{2}at^2$

with x replaced by y. 

$0=0+v_{0y}(3.4 \ s)+\frac{1}{2}(-9.8 \ ms/^2)(3.4 \ s)^2$

$v_0=16.66 \ m/s$

The height can be calculated from Eq. 

$v_y^2=v_{0y}^2+2a(y-y_0)$

with a final velocity of $v_y$ = 0 at the top of the path. 

$0^2=(16.66 \ m/s^2)^2+2(-9.80)(y-0)$

$y=\frac{-(16.66 \ m/s)2}{-19.6 \ m/s^2}$

y = 14 m

Q#40

(II) A baseball is hit almost straight up into the air with a speed of 25 m/s. Estimate 

(a) how high it goes, 

(b) how long it is in the air. 

(c) What factors make this an estimate?

Answer:

Choose upward to be the positive direction, and take 0y = 0 to be at the height where the ball was hit. For the upward path, $v_{0y}$= 25 m/s, $v_y$ = 0 at the top of the path, and a = −9.80 $m/s^2$.

(a) The displacement can be found from Eq. 

$v_y^2=v_{0y}^2+2a(y-y_0)$, with x replaced by y.

$0^2=(25.0 \ m/s^2)^2+2(-9.80)(y-0)$

$y=\frac{-(25.0 \ m/s)2}{-19.6 \ m/s^2}$

y = 32 m

(b) The time of flight can be found from Eq. 

$y=y_0+v_{0y}t+\frac{1}{2}at^2$

with x replaced by y, using a displacement of 0 for the displacement of the ball returning to the height from which it was hit. 

$0=0+v_{0y}t+\frac{1}{2}at^2$

$0=t(v_{0y}t+\frac{1}{2}at)$ so

t = 0 or $t=\frac{2v_{0y}}{-a}=\frac{2(25.0 \ m/s)}{-(-9.80 / m/s^2)}=5.1 \ s$

The result of t = 0 s is the time for the original displacement of zero (when the ball was hit), and the result of t = 5.1 s is the time to return to the original displacement. Thus the answer is t = 5.1 seconds.

 (c) This is an estimate primarily because the effects of the air have been ignored. There is a nontrivial amount of air effect on a baseball as it moves through the air—that’s why pitches like the “curve ball” work, for example. So ignoring the effects of air makes this an estimate. Another effect is that the problem says “almost” straight up, but the problem was solved as if the initial velocity was perfectly upward. Finally, we assume that the ball was caught at the same height as which it was hit. That was not stated in the problem either, so that is an estimate. 

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