Question

a body is projected with a speed u m/* at an angle b with the horizontal . the k.e at the highest pt. is 3/4 th of the initial k.e . the value of b is

Answer 1

As there is no acceleration in horizontal direction the velocity component will remain same as VcosƟ and at highest point vertical component will be zero.

So total velocity at top will be VcosƟ

Kinetic energy at the highest point will be= m( VcosƟ)2

Initial kinetic energy will be =1/2mV2

At highest point

kE at highest =3/4 kE initial

1/2 m( VcosƟ)2 =1/2mV2 *3/4 cosƟ=(3) 12/2

Ɵ=30 degree