A ball thrown upwards takes 4 s to reach the maximum height. Find (a) the initial speed with which it was thrown, and (b) the maximum height reached.

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Solution

$I n i t i a l s p e e d i s u m a x i m u m h e i g h t i s h f i n a l v e l o c i t y a t t o p m o s t p o i n t i s z e r o v = 0 t i m e t a k e n t = 4 s v = u + a t a c c e l e r a t i o n a = - g 0 = u - 4 g u = 4 g u = 4 \times 10 = 40 m / s h = u t + 0.5 a t^{2} h = 40 \times 4 - 0.5 \times 10 \times 16 h = 80 m$

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A ball thrown upwards takes 4 s to reach the maximum height. Find (a) the initial speed with which it was thrown, and (b) the maximum height reached.

Initial speed is umaximum height is hfinal velocity at top most point is zero v = 0time taken t = 4 sv = u + atacceleration a = -g0 = u - 4gu = 4gu = 4×10 = 40 m/sh = ut + 0.5at2h = 40×4-0.5×10×16h = 80 m