First slide

Projection Under uniform Acceleration

Question

The maximum height reached by projectile is 4 m. The horizontal range is 12 m. The velocity of projection in ms^-1 is (g is acceleration due to gravity)

Moderate

A

$5 \sqrt{g / 2}$
B

$3 \sqrt{g / 2}$
C

$\frac{1}{3} \sqrt{g / 2}$
D

$\frac{1}{5} \sqrt{g / 2}$

Solution

$\tan θ = \frac{4 H}{R} = \frac{4 \times 4}{12} = \frac{4}{3} \Rightarrow \sin θ = \frac{4}{5} H = \frac{u^{2} \sin^{2} θ}{2 g} \Rightarrow u = \frac{\sqrt{2 gH}}{\sin θ} = \frac{\sqrt{2 \times g \times 4}}{4 / 5} = 5 \sqrt{\frac{g}{2}}$

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