In the question number 67, the time taken by the ball to reach the ground is

2 s

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3 s

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5 s

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7 s

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Solution

The correct option is C 5 s
Taking upward direction as $+ v e,$
$u = + 20 m / s a = - 10 m / s^{2} s = - 25 m [To the point on the ground.]$
By third law of motion,
$v^{2} = u^{2} + 2 a s v^{2} = {(20)}^{2} + 2 (- 10) (- 25) v^{2} = 400 + 500 v = 30 m / s (downwards)$
Now, by first equation of motion,
$v = u + a t - 30 = 20 - 10 t t = 5 s e c$

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