First slide

Projection Under uniform Acceleration

Question

The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be

Moderate

A

50 m
B

100 m
C

25 m
D

37 m

Solution

$\begin{array}{l} R_{1} = \frac{u^{2} \sin 2 θ_{1}}{g}; R_{2} = \frac{u^{2} \sin 2 θ_{2}}{g} \\ \frac{R_{2}}{R_{1}} = \frac{\sin 2 θ_{2}}{\sin 2 θ_{1}} = \frac{\sin (2 {.45}^{\circ})}{\sin (2 {.15}^{\circ})} \\ = \frac{\sin 90^{\circ}}{\sin 30^{\circ}} = \frac{1}{1 / 2} = 2 \end{array}$
R₂ = 2 R₁ = 2. 50 = 100 m.

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