Question

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

Answer 1

Kinetic energy of projection
= 0.5mu²

Potential energy at maximum height
= mgh
= mg × (u²sin²θ) / (2g)
= (mu²sin²θ) / 2

According to question
(mu²sin²θ) / 2 = (3/4) × 0.5mu²
Divide both sides by mu²
sin²θ / 2 = 3/8
sin²θ = 3 / 4
sinθ = √3 / 2
sinθ = sin60
θ = 60°

Angle of projection is 60°

Answer 2

ANGLE OF PROJECTION IS 60°.

Given:

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

To find:

Angle of projection ?

Calculation:

Let the angle of projection be $\theta$

• At max height, PE will be mgh
• At projection, KE = ½mu²

$PE_{max \: ht.} = \dfrac{3}{4} \times KE_{projection}$

$\implies mgh= \dfrac{3}{4} \times \dfrac{1}{2} m {u}^{2}$

$\implies mg \times \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g} = \dfrac{3}{4} \times \dfrac{1}{2} m {u}^{2}$

$\implies \dfrac{ m{u}^{2} { \sin}^{2}( \theta) }{2} = \dfrac{3}{4} \times \dfrac{1}{2} m {u}^{2}$

$\implies { \sin}^{2}( \theta) = \dfrac{3}{4}$

$\implies \sin( \theta) = \dfrac{ \sqrt{3} }{2}$

$\implies \theta = {60}^{ \circ}$

So, the angle of projection is 60°.