A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of $10 m s^{- 2}$ , with what velocity will it strike the ground and time after which it will it strike the ground?

m / s

, 2 s

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m / s

\sqrt{2}

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m / s

, 4 s

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None of the above

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Solution

The correct option is A 20 $m / s$ , 2 s
Given, initial velocity of ball, $u = 0$
Final velocity of ball, $v = ?$
Distance through which the balls falls, $s = 20 m$
Acceleration $a = 10 m s^{- 2}$
Time of fall, $t = ?$
We know
$v^{2} - u^{2} = 2 a s$
or $v^{2} - 0 = 2 \times 10 \times 20 = 400$ or $v = 20 m s^{- 1}$
Now using $v = u + a t$ we have
$20 = 0 + 10 \times t$ or $t = 2 s$

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A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms−2, with what velocity will it strike the ground and time after which it will it strike the ground?

The correct option is A 20 m/s, 2 sGiven, initial velocity of ball, u=0Final velocity of ball, v=?Distance through which the balls falls, s=20mAcceleration a=10ms−2Time of fall, t=?We knowv2−u2=2asor v2−0=2×10×20=400 or v=20ms−1Now using v=u+at we have20=0+10×t or t=2s

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of $10 m s^{- 2}$ , with what velocity will it strike the ground and time after which it will it strike the ground?