Question

A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is $α$ = 30°, the angle of the barrel to the horizontal $β$ = 60°. The initial velocity v of the shell is 21 m/s. Then, the distance of point from the gun at which shell falls is

Moderate

Solution

Here, u = 21 m/s, $α = 30^{\circ}, θ = (β - α)$
$= 60^{\circ} - 30^{\circ} = 30^{\circ}$
Maximum range
$\begin{array}{l} R = \frac{2 u^{2}}{g} \frac{\sin θcos (θ + α)}{\cos^{2} α} \\ = \frac{2 \times (21)^{2} \times \sin 30^{\circ} \cos 60^{\circ}}{9 .8 \times \cos^{2} 30^{\circ}} = 30 m \end{array}$

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