A body is dropped from a height of 39.2 m. After it crosses half the distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take g=9.8 m/ $s^{2}$ )

19.6 m/s

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28 m/s

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1.96 m/s

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196 m/s

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Solution

The correct option is A
19.6 m/s

Suppose v be the velocity of the body after falling through half the distance. Then

s = $\frac{39.2}{2}$ = 19.6 m , u = 0 and g = 9.8 m/ $s^{2}$

$v^{2}$ = $u^{2}$ + 2gs = $0^{2}$ + 2 $\times$ 9.8 $\times$ 19.6

V = 19.6 m/s

Then, it continues to move with same velocity till it hits the ground.

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A body is dropped from a height of 39.2 m. After it crosses half the distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take g=9.8 m/s2 )

The correct option is A 19.6 m/s Suppose v be the velocity of the body after falling through half the distance. Then s = 39.22 = 19.6 m , u = 0 and g = 9.8 m/ s2 v2 = u2 + 2gs = 02 + 2 × 9.8 × 19.6 V = 19.6 m/s Then, it continues to move with same velocity till it hits the ground.

A body is dropped from a height of 39.2 m. After it crosses half the distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take g=9.8 m/ $s^{2}$ )