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 (let than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second balls is +15 m at t = 2s. 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 two balls (1 & 2) be v₁ and v₂, if v₁ = 2v, v₂ = v

If y₁ and y₂ the displacement covered by the balls 1 and 2, respectively, then

 \(y_1=\frac{v^2_1}{2g}=\frac{4v^2}{2g}\) and \(v_2=\frac{v^2_2}{2g}=\frac{v^2}{2g}\)

since, y₁ − y₂ = 15 m,

\(\frac{4v^2}{2g}-\frac{v^2}{2g}\) = 15 m

or \(\frac{3v^2}{2g}\) = 15 m

\(v^2=\sqrt{5m\times(2\times10)}\) m/s²

v = 10 m/s

clearly, v₁ = 20 m/s, v₂ = 10 m/s.

Time interval = 1 s.