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)

Moderate

A

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

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

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

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

Solution

$R = 12 m \frac{H}{R} = \frac{1}{4} tanθ$
$H = 4 m \frac{4}{12} = \frac{1}{4} tanθ$
$tanθ = \frac{4}{3} sinθ = \frac{4}{5}$
$H = \frac{u^{2} \sin^{2} θ}{2 g} H = u^{2} \frac{16}{25 \times 2 g}$
$u^{2} = \frac{100 \times 2 g}{16}$
$u = 5 \sqrt{g / 2}$

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