A ball is thrown vertically upwards from the edge of a rooftop of a tall building. It passes a point 50 m below the rooftop after 5 s it was thrown.

Let ball A is thrown upward at t = 0. It takes time T to reach the ground. Ball B is dropped from the roof 1.00 s later. If both the balls reach the ground simultaneously, the flying time for ball B will be (T- 1) s.

The displacement of both the balls is (= -20 m).

Using y = y0 + ut + (1/2)at2 initially for both balls are at y0 = 20 and when both reaches ground where y = 0.

For ball A: 0 = 20 + v0T + (1/2)(-10)T2     ...(i)

For ball B: 0 = 20 + (1/2)(-10)(T−1)2     ...(ii)

From Eqs. (i) and (ii),

v0T = 5(2T - 1)     ...(iii)

From Eqs. (i) and (iii), we get

T2 - 2T - 3 = 0 ⇒ (T + 1)(T - 3) = 0

T = 3s

From Eqs.(iii) v0 = 25/3 m/s

Let ball A reaches maximum height at time t' after throwing. For ball A at maximum height, the velocity will be zero.

Using v = u + at

0 =(25/3) - 10 ×t′⇒t′=5/6 s

For maximum height reached by ball A,

Using v2=u2+2as

0 = (325)2 - 2 x 10 x h'

h' = 125/36 m

Hence, maximum height reach by ball A

H = 20 + (125/36) = 845/36 m

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