Question

OBJECTIVE PHYSICS-XIAS
The potential energy of a projectile at maximum height is 3/4 times kinetic energy of projection. Its
angle of projection is
2) 45°
3) 60°
4) none
1) 30°
​

Answer 1

Answer:

bsuuhsh

Explanation:

n

$.41? \times \frac{?}{?} \times \frac{?}{?}$

Answer 2

Answer:

Option (3) 60°

Explanation:

If the angle of projection is θ and the initial velocity u then the maximum height of the projectile

$H=\frac{u^2\sin^2\theta}{2g}$

If m is the mass of the object, then

The potential energy at maximum heigh = mgH

Kinetic Energy of projection = $\frac{1}{2}mu^2$

According to the question

$mgH=\frac{3}{4}\times \frac{1}{2}mu^2$

$\implies gH=\frac{3}{4}\times \frac{1}{2}u^2$

$\implies g\frac{u^2\sin^2\theta}{2g}=\frac{3}{4}\times \frac{1}{2}u^2$

$\implies \sin^2\theta=\frac{3}{4}$

$\implies \sin\theta=\frac{\sqrt{3}}{2}$

$\implies \sin\theta=\sin60^\circ$

$\implies \theta=60^\circ$

Hope the answer is helpful.