Physics, asked by vraj82, 10 months ago

a stone is released from the top of a tower of height 200m.find the distance covered by the stone during the fifth second of it's motion.

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Answered by ferozemulani

15

Explanation:

distance travelled in 5th sec =

u + (g/2)*(2*5 -1)

= 0 + 4.9*(9) = 44.1 m

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