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?

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° (B) 60° (C) 45° (D) 120°

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/64ac3361e54cf5d444b8d1e42c058919-.png" /> According to the given problem (1/2) m(u cos θ)2=(3/4) × (1/2) m u2 or cos 2 θ=(3/4) or cos θ=(√3/2)= cos 30° or θ=30°

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

$3 0^{\circ}$

$6 0^{\circ}$

$4 5^{\circ}$

$12 0^{\circ}$

According to the given problem
$\frac{1}{2} m (u cos θ)^{2} = \frac{3}{4} \times \frac{1}{2} m u^{2}$
or $cos^{2} θ = \frac{3}{4}$
or $cos θ = \frac{3}{2} = cos 3 0^{\circ}$ or $θ = 3 0^{\circ}$

Q. A body is projected such that its kinetic energy at the top is (3/4)th of its initial kinetic energy. What is the angle of projection with the horizontal?

30∘

60∘

45∘

120∘