First slide

Projectile motion

Question

The speed of a projectile at its maximum height is $\sqrt{3} / 2$ times its initial speed. If the range of the projectile is P times the maximum height attained by it, P is equal to

Moderate

A

$4 / 3$
B

$2 \sqrt{3}$
C

$4 \sqrt{3}$
D

$3 / 4$

Solution

$Given \frac{\sqrt{3} u}{2} = u \cos θ = speed at maximum height$
$or \cos θ = \frac{\sqrt{3}}{2} or θ = 30^{\circ} . . . . . . . (i)$
$Given that P H_{max} = R . . . . . . (i i)$
$We know H_{max} = \frac{R \tan θ}{4}$
$P = \frac{R}{H_{max}} = \frac{4}{\tan θ} = \frac{4}{\tan 30^{\circ}} = 4 \sqrt{3}$

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