Question

You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 41.1 m. (a) At what time do they pass each other? s (b) At what location do they pass each other relative to the window?

Answer 1

Answer:1.44 s,10.17 m

Explanation:

Given

two balls are separated by a distance of 41.1 m

One person drops the ball from a height of 41.1 and another launches a ball with velocity of 41.1 m exactly at the same time.

Let the ball launches from ground travels a distance of x m in t sec

For Person on window

$41.1-x=ut+\frac{1}{2}gt^2$

$41.1-x=0+\frac{1}{2}\times 9.81\times t^2--------1$

For person at ground

$x=v_ft-\frac{1}{2}gt^2---------2$

add (1) & (2)

$41.1=v_ft-\frac{1}{2}gt^2+\frac{1}{2}gt^2$

$41.1=v_ft$

and $v_f$ is given by

$v_f=\sqrt{2\times 9.81\times 41.1}=28.39 m/s$

$t=\frac{41.1}{28.39}=1.44 s$

Substitute value of t in equation 1

$41.1-x=0+\frac{1}{2}\times 9.81\times 1.44^2$

41.1-x=10.171 m

Thus the two ball meet at distance of 10.17 m below the window.