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13. A ball is dropped from certain height such that thedistance travelled in last second is equal to thedistance travelled in first three second. Find its height​

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Formula USED:

sn=u+g/2(2n−1)

Answer:125m

Explanation:

The distance covered in first 3 second

s=1/2gt^2

=1^2×10×3^2

=45

Ball TAKES 'n' seconds to fall to the GROUND. The distance covered in the NTH second is

sn=u+g/2(2n−1)

=10/2(2n−1)

=10n−5

Therefore 45=10n−5

n=5=t

Therefore h=12*gt^2

=12×10×25

=125m



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