Question

78. The velocity at the maximum height of a proj-
ectile is half of its initial velocity of projection.
The angle of projection is
1) 30° 2) 45° 3) 60°

4) 76°​

Answer 1

Answer:

$\boxed{\mathfrak{(3) \ 60^{\circ}}}$

Explanation:

Let initial velocity of projectile be $\sf v_0$ making angle $\sf \theta$ with the horizontal.

Velocity of projectile at maximum height is equal to x-component of initial velocity i.e. $\sf v_0 cos \theta$

According to the question,

Velocity at maximum height of the projectile is half of its initial velocity of projection.

$\sf \implies \cancel{v_0} cos \theta = \frac{ \cancel{v_0}}{2} \\ \\ \sf \implies cos \theta = \frac{1}{2} \\ \\ \sf \implies \theta = {cos}^{ - 1} ( \frac{1}{2} ) \\ \\ \sf \implies \theta = {60}^{ \circ}$

$\therefore$

Angle of projection = 60°