Question

A body is projected such that its kinetic energy at the top is $(3 / 4)^{\text {th }}$ of its initial kinetic energy. What is the angle of projection with the horizontal?

• A. $30^{\circ}$
• B. $60^{\circ}$
• C. $45^{\circ}$
• D. $120^{\circ}$

Answer 1

Answer: (A)

According to the given problem
$\frac{1}{2} m(u \cos \theta)^{2}=\frac{3}{4} \times \frac{1}{2} m u^{2}$
or $\cos ^{2} \theta=\frac{3}{4}$
or $\cos \theta=\frac{\sqrt{3}}{2}=\cos 30^{\circ}$ or $\theta=30^{\circ}$