Question

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 2 seconds). 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 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.

Answer 1

Let the speeds of the two balls (1 and 2) be v₁ and v₂ where

If v₁ = 2v, v₂ = v

If y₁ and y₂ and the distance covered by balls 1 and 2, respectively before coming to rest then

`y_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)` = 15 m or `(3v^2)/(2g)` = 15 m

or `v^2 = sqrt(5m xx (2 xx 10))` m/s²

or v = 10 m/s

Clearly, v₁ = 20 m/s and v₂ = 10 m/s

As `y_1 = v_1^2/(2g) = (20 m)^2/(2 xx 10  m 15)` = 20 m

`y_2 - y_1 - 15` m = 5 m

It t₂ is the time taken by the ball 2 toner a distance of 5 m, then from `y_2 = v_2^t - 1/2 "gt"_2^2`

5 = `10t_2 - 5t_2^2` or `t_2^2 - 2t_2 + 1` = 0

Where t₂ = 15

 Since t₁ (time taken by ball 1 to cover the distance of 20 m) is 2s, the time interval between the two throws

= t₁ – t₂

= 2s – 1s

= 1s