First slide

Projection Under uniform Acceleration

Question

A shell fired from the ground is just able to cross horizontally the top of a wall 90 m away and 45 m high. The direction of projection of the shell will be

Easy

A

25°
B

30°
C

60°
D

45°

Solution

$h = \frac{u^{2} \sin^{2} θ}{2 g} and R = \frac{u^{2} \sin 2 θ}{g} \frac{h}{R} = \frac{45}{180} = \frac{\tan θ}{4} or θ = 45^{\circ}$

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