Question

what is ratio of P.E. wr.t. ground and K.E. at the top most point of projectile motion:-​

Answer 1

Answer:

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Answer 2

Explanation:

If I throw a projectile with speed u at an angle Q with ground, then

H = (u sinQ)^2/2g

Velocity at top most point = v=ucosQ.

P.E at top most point = mgH =m(u sinQ)^2/2

K.E at top most point = mv^2/2 =m(u cosQ)^2/2

I calculated the ratio = (tanQ)^2