Question

- A body is thrown vertically upward with velocity u. The
ratio of times at which it is at particular height h is 1 : 2
What is the maximum height reached by the body above
point of projection
(a) (9/4)h
(b) (4/3)h
(c) (9/8)h
(d) none of these​

Answer 1

Answer:

I think A is the right answer

Answer 2

Answer:

Hii mate GM ❤

Explanation:

☸Your solution

Le H is the maximum height. Let h is the height above the ground at which both P.E and K.E are equal

∴ T e at the point p = K.E + P.E

⇒ K.E = mgH - mgh = mg [H - h]

∴ P.E / K.E = mgh/mg[Hh]

As K.E = P.E

∴ h / H - h = 1, H – h = 2h …(1)

When the body is projected upwards with velocity u it reaches a maximum height H

∴ H = u2/ 2g

When the body is projected vertically upwards with double the velocity then it reaches a maximum height H1

∴ H = (2u)2/2g = 4u2/2g = 4H………..(2)

∴ The ratio of P.E and K.E at the same point ‘P’ = P.E / K.E = mgh/mg[H1-h]

∴ P.E / K.E = h/4[2h]-h

= 1/7

Or, 1:7

Hope this helps you buddy :)