First slide

Projection Under uniform Acceleration

Question

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

Moderate

A

$\frac{4}{3}$
B

$2 \sqrt{3}$
C

$4 \sqrt{3}$
D

$\frac{3}{4}$

Solution

Given $\frac{\sqrt{3}}{2} u = u \cos θ = speed at maximum height$
$\cos θ = \frac{\sqrt{3}}{2}$
or $θ = 30^{0}$ ------------------------(i)
Given that ${nH}_{\max .} = R - - - - - (ii)$
We know $H_{\max .} = \frac{R \tan θ}{4}$
$n = \frac{4}{t a n θ} = \frac{4}{t a n 30^{0}} = 4 \sqrt{3}$

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