Answer to Question #100857 in Classical Mechanics for Sridhar

Question

A ball is projected vertically up from ground.Boy 'A' Standing at the window of first floor of a nearby building observes that the time interval between the ball crossing him while going up and the ball crossing him while going down is 2 s. Another boy 'B' standing on the second floor notices that time interval between the ball passing him twice during up motion and down motion is 1 s.Calculate the difference between the vertical positions of boy B and boy A.
(Assume g=10m/s^(2))
Ans: 3.75m

Answer 1

Let $v_0$ is an initial speed of the ball. The total time of motion of the ball is 2 sec, so initial velocity

$$v_0=gt/2=10\times 2/2=10\:\rm m/s.$$

The time of motion of the ball between two boys is $\tau=0.5$ sec. Hence the difference between the vertical positions of boy B and boy A

$$d=v_0\tau-g\tau^2/2$$ $$=10\times 0.5-10\times 0.5^2/2=3.75\:\rm m$$