A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2s). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is 15 m at t = 2. The gap is found to remain constant. The velocities with which the balls were thrown are (Take g = `10 m s^(-1)`)

Options

A. `20 m s^(-1) , 10 m s^(-1)`
B. `10 m s^(-1) , 5 m s^(-1)`
C. `16 m s^(-1) , 8 m s^(-1)`
D. `30 m s^(-1), 15 m s^(-1)`

Answer

A

Explanation

For first stone,
taking the vertical upwards motion of the first stone up to highest point
Here, u = `u_1`, v = 0 (At highest point velocity is zero)
a = -g, S = `h_1`
As `v^(2) - u^(2) = 2aS` `therefore (0)^(2) - u_(1)^(2) = 2(-g)h_1` or `h_1 = u_(1)^(2)/(2g)` ...(i)
For second stone,
Taking the vertical upwards motion of the second stone up to highest point
here, u = `U_2, v = 0, a = -g, S = h_2`
As `v^(2) - u^(2)` = 2as
`therefore (0)^(2) - (u_2)^(2) = 2(-g)h_2` or `h_2 = u_(1)^(2)/(2g)`.............(ii)
As per question
`H_1 - h_2 = 15 m , u_2 = u_1/2`
Subtract (ii) from (i), we get, `h_1 - h_2 = u_(1)^(2) /(2g)-(u_(2)^(2)/(2g))`
On substituting the given information, we get
`15 = u_(1)^(2)/(2g)-u_(1)^(2)/(2g)=(3u_(1)^(2))/(8g)` or `u_(1)^(2)=(15 xx 8g)/(3)=(15 xx 8 xx 10)/(3) = 400`
or `u_1 = 20 m s^(-1)` and `u_2 = U_1/2 = 10 m s^(-1)`