Question

The potential energy of a projectile at maximum height is 3/4 times kinetic energy of projection. Its angle of projection is

Answer 1

answer
angle of projection
= v/ u root 1 / 6 sin ^ -1

solution
3/ 4 mgh = 1/2 mv^2
gh = v^2 / 3
h = v^2 / 3g

at maximum height :
h = u^2 sin^2 theta / 2g

v^ 2 / 3g = u^2 sin^2 theta / 2g

Sin^2 theta = v^2 / 6 u^2
Sin theta = v/ u x root 1/ 6
theta = v/u root 1/6 Sin^-1
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