MCQ
A projectile is thrown with some initial velocity at an angle $\alpha$ to the horizontal. Its velocity when it is at the highest point is $(2 / 5)^{1 / 2}$ times the velocity when it is at height half of the maximum height. Find the angle of projection $\alpha$ with the horizontal.
  • A
    $30$
  • B
    $45$
  • $60$
  • D
    $37$

Answer

Correct option: C.
$60$
c
(c)

$(u \cos \alpha)=\sqrt{\frac{2}{5}} \sqrt{\left(u \cos \alpha^2\right)+\left\{(u \sin \alpha)^2-2 g h\right\}}$

Here, $h=\frac{H}{2}=\frac{u^2 \sin ^2 \alpha}{4 g}$

Solving this equation, we get $\alpha=60^{\circ}$

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