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 and second ball is `+15m`.
Q. The speed of first ball is

Options

A. 20 m/s
B. 10 m/s
C. 5 m/s
D. 15 m/s

Answer

A

Explanation

Let the speeds of the two balls (1 and 2) be `v_1` and `v_2` where `v_1=2v`,`v_2=v`
If `y_1` and `y_2` and the distance covered by the balls 1 and 2, respectively before coming to rest, then
`t_1=(v_1^2)/(2g)=(4v^2)/(2g)` and `y_2=(v_2^2)/(2g)=(v^2)/(2g)`
Since `y_1-y_2=15`m,`(4v^2)/(2g)-(v^2)/(2g)`
`(4v^2)/(2g)-(v^2)/(2g)=15`
`impliesv=sqrt(5xx(2xx10))=10(m)/(s)`