First slide

Projection Under uniform Acceleration

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

A

10 m
B

20 m
C

30 m
D

40 m

Solution

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

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