Question

A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50 m tall, you observe that it takes the ball 0.150 s to traverse the length of the window. 1) Determine how high above the top of your window the ball was dropped. Ignore the effects of air resistance. (Express your answer to three significant figures.)

Answer 1

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

Given:

• Height of the window=1.5 m
• Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

$s=ut+\dfrac{at^2}{2}$

Let u be the velocity of the ball when it reaches the top of the window

then

$1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s$

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

$3.96=gt\\t=0.4\ \rm s$

Let h be the height above the top of the window

$h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m$