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:

9.8 m

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49 m

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24.5 m

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19.6 m

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Solution

The correct option is C $24.5 m$
Time for $1^{s t}$ ball $= 2 s$
Time for $2^{n d}$ ball $= 2 + 1 = 3 s$
$s_{1} = \frac{1}{2} g (2)^{2} a n d s_{2} = \frac{1}{2} g (3)^{2}$
$s_{2} - s_{1} = g / 2 (9 - 4) = 4.9 \times 5 = 24.5 m$

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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).

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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 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:

The correct option is C 24.5mTime for 1st ball =2sTime for 2nd ball =2+1=3ss1=12g(2)2ands2=12g(3)2s2−s1=g/2(9−4)=4.9×5=24.5m

The correct option is C $24.5 m$
Time for $1^{s t}$ ball $= 2 s$
Time for $2^{n d}$ ball $= 2 + 1 = 3 s$
$s_{1} = \frac{1}{2} g (2)^{2} a n d s_{2} = \frac{1}{2} g (3)^{2}$
$s_{2} - s_{1} = g / 2 (9 - 4) = 4.9 \times 5 = 24.5 m$