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A ball is dropped from the top of the tower Of height h. It covers a distance of h/2 in the lasy second of its motion. How long does the ball remains in the air?

A ball is dropped from the top of the tower Of height h. It covers a distance of h/2 in the lasy second of its motion. How long does the ball remains in the air?

Grade:11

1 Answers

Arun
25750 Points
5 years ago

Let the time during which the ball remains in air be T seconds

Then we have h=1/2 gT^2 …(1)

Now distance travelled by ball in nth second is given by Sn=g/2(2n-1)

h/2=g/2(2T-1)…(2)

Substituting h from (1) in (2) we get

T^2=4T-2

T^2–4T+2=0

Which gives T=3.41 seconds

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