From the top of a building of height $40 m$ , a boy throws a stone vertically upwards with an initial velocity of $10 m s^{- 1}$ such that it eventually falls to the ground. At what time will the stone reach the ground? $(g = 10 m s^{- 2})$

4 s

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3 s

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2 s

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1 s

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Solution

The correct option is A $4 s$
$s = - 40 m, u = + 10 m / s, a = - 10 m / s^{2}$
Now $s = u t + (1 / 2) a t^{2}$
=> $- 40 = 10 t + 1 / 2 \times (- 10) t^{2}$
=> $t^{2} - 2 t - 8 = 0$
=> $(t + 2) (t - 4) = 0$
=> $t = - 2, 4 s$
The negative value of t is not possible
Hence the ball will hit the ground after $4 s$

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