A ball is thrown vertically upwards with a velocity of 20 ms -1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25 m from the ground. The height to which the ball rises from the ground and the time taken before the ball hits the ground are (Take g = 10 ms -2 )

Question

A ball is thrown vertically upwards with a velocity of 20 ms -1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25 m from the ground. The height to which the ball rises from the ground and the time taken before the ball hits the ground are (Take g = 10 ms -2 ) (A) 20 m , 2 s (B) 40 m , 3 s (C) 40 m , 5 s (D) 45 m , 5 s

Tardigrade Admin · Accepted Answer

Taking point of launch as origin and up as + ve, using

v^{2} = u^{2} + 2 a s

0^{2} = (20 m s^{- 1})^{2} + 2 (- 10 m s^{- 2}) \cdot H_{m a x}

\Rightarrow H_{m a x} = \frac{2 0 ^{2}}{20} = 20 m

Hence, from the ground the maximum height reached

= 25 m + 20 m = 45 m

Further, for time to reach ground

s = - 25 m, u = + 20 m s^{- 1}, a = - 10 m s^{- 2}

Using,

s = u t + \frac{1}{2} a t^{2} \Rightarrow - 25 = 20 t - 5 t^{2}

5 t^{2} - 20 t - 25 = 0

or

5 t^{2} - 25 t + 5 t - 25 = 0

5 t (t - 5) + 5 (t - 5) = 0 \Rightarrow (5 t + 5) \cdot (t - 5) = 0

t = - 1 s

or

t = 5 s

Rejecting -ve value of time, we get

t = 5 s

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