A man is standing on top of a building 100 m high. He throws two balls

1.	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» 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.

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»

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.