First slide

Projection Under uniform Acceleration

Question

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

Easy

A

$\sqrt[5]{\frac{g}{2}}$
B

$\sqrt[3]{\frac{g}{2}}$
C

$\frac{1}{3} \sqrt{\frac{8}{2}}$
D

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

Solution

$4 H = R \tan θ \Rightarrow 4 \times 4 = 12 \times \tan θ \Rightarrow \tan θ = \frac{4}{3}$
Given $H_{max} = \frac{u^{2} \sin^{2} θ}{2 g} = 4$
$\begin{array}{l} \Rightarrow u^{2} \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{2 \times 10} = 5 \\ u^{2} = 125 \Rightarrow u = 5 \sqrt{5} m = 5 \sqrt{\frac{g}{2}} \end{array}$

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