Question

In the case of a projectile the K.E and P.E are equal at the maximum height attained. The angle of projection is;

Answer 1

Answer:

K.E.+P.E.=U

P.E=-1/K.E.

Answer 2

Answer:

At max height velocity  will be only in horizontal direction ..

Explanation:

PE=KE

$mg\frac{v^{2}sin^{2}\theta }{2g} =\frac{1}{2}m(vcos\theta)^{2} \\(\frac{vsin\theta}{vcos\theta} )^{2} =1\\tan\theta=1\\\theta=45$