A man is standing on top of a building 100 m high. He throws two ball vertically, one at `t=0` and after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. At `t=2`, both the balls reach to their maximum heights at this time the vertical gap between first and second ball is `+15m`.
Q. The speed of first ball is

Options

A. 1.2 sec
B. 0.5 sec
C. 0.8 sec
D. 1 sec

Answer

D

Explanation

As `y_1=(v_1^2)/(2g)=((20)^2)/(2xx10)=20`m
`y_2=y_1-15m=5m`.
it `t_2` is the time taken by the ball 2 to cover
a distance of 5 m then from `y_2=v_2t-(1)/(2)gt_2^2`
`5=10t_2-5t_2^2` or `t_2^2-2t_2+1=0`
Where `impliest_2=1`sec
Since `t_1(time taken by ball 1 to cover distance of 20m) is 2s. Time interval between the two throws
`=t_1-t_2=2s-1s=1s`.