First slide

Projection Under uniform Acceleration

Question

A particle is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

Moderate

A

$\frac{4 v^{2}}{5 g}$
B

$\frac{4 g}{5 V^{2}}$
C

$\frac{4 v^{3}}{5 g^{2}}$
D

$\frac{4 v}{5 g^{2}}$

Solution

Since R = 1H
$\frac{v^{2} \sin 2 θ}{g} = 2 \times \frac{v^{2} \sin^{2} θ}{2 g} 2 sinθcosθ = \sin^{2} θ or tan θ = 2 R = \frac{v^{2} \sin 2 θ}{g} = \frac{v^{2} 2 sinθcosθ}{g} = \frac{2 v^{2}}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} = \frac{4 v^{2}}{5 g}$

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