First slide

Projection Under uniform Acceleration

Question

The angle of projection, of which the horizontal range and maximum height of projectile are equal is

Moderate

A

$45^{\circ}$
B

$θ = \tan^{- 1} (0 \cdot 25)$
C

$θ = \tan^{- 1} 4$
D

$60^{\circ}$

Solution

We know that
$R = \frac{u^{2} \sin 2 θ}{g} and h_{max} = \frac{u^{2} \sin^{2} θ}{2 g}$
Given that $R = h_{max}$
$\begin{array}{l} ∴ \frac{u^{2} \sin 2 θ}{g} = \frac{u^{2} \sin^{2} θ}{2 g} \\ or 2 \sin θcos θ = (\sin^{2} θ / 2) \\ or \tan θ = 4 or θ = \tan^{- 1} (4) \end{array}$

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