A body is released from a great height falls freely towards the earth. Another body is released from the same height exactly a second later. Then, the separation between the two bodies, 2 sec after the release of the second body is nearly:

15 m

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20 m

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25 m

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30 m

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Solution

The correct option is C 25 m
$S = u t + \frac{1}{2} g t^{2} S = 5 t^{2} S_{1} = (t = 3 s) = 5 \times 3^{2} = 45 S_{2} = (t = 2 s) = 5 \times 2^{2} = 20 S_{1} - S_{2} = 45 - 20 = 25 m$

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A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is

Q. A body released from a height falls freely towards earth. Another body is released from the same height exactly one second later. The seperation between the two bodies two second after the release of the second body is:

A body released from a height falls freely towards earth. Another body is released one sec later. The separation between the bodies two sec after the released of the second body is (ans=24.5 m).

A body released from a great height falls freely towards the earth.Another body is released from the same height exactly one second later. The separation between both the bodies two seconds after the release of the second body is

a) 24.5 metre

b)34.5 metre

c) 44.5 metre

d) 54.5 metre

Q. A body is released from a height and it falls freely towards the earth. Exactly 1 s later another body is released. What is the distance between the two bodies 2 s after the release of the second body?

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A body is released from a great height falls freely towards the earth. Another body is released from the same height exactly a second later. Then, the separation between the two bodies, 2 sec after the release of the second body is nearly:

The correct option is C 25 mS=ut+12gt2S=5t2S1=(t=3s)=5×32=45S2=(t=2s)=5×22=20S1−S2=45−20=25 m

The correct option is C 25 m
$S = u t + \frac{1}{2} g t^{2} S = 5 t^{2} S_{1} = (t = 3 s) = 5 \times 3^{2} = 45 S_{2} = (t = 2 s) = 5 \times 2^{2} = 20 S_{1} - S_{2} = 45 - 20 = 25 m$