Question

The potential energy of a projectile at its highest point is (1/2)th the value of its initial kinetic energy. Therefore its angle of projection is ...... 

(A) 30° 

(B) 45° 

(C) 60° 

(D) 75°

Answer 1

Correct option: (B) 45°

Explanation:

Given: PE at h_max = (1/2) × initial KE

mgh_max = (1/2) × (1/2) mv² 

gh_max = (1/4)V² 

h_max for projectile = {(V² sin² θ) / (2g)}

g {(V² sin² θ) / (2g)} = (1/4) V² 

sin²θ = (1/2)

sin θ = (1 / √2)

θ = 45°