First slide

Projection Under uniform Acceleration

Question

For angles of projection of a projectile at angles $(45^{\circ} - θ) and (45^{\circ} + θ)$ , the horizontal range described by the projectile are in the ratio of

Easy

A

2 : 1
B

1 : 1
C

2 : 3
D

1 : 2

Solution

Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g}$
For angle of projection $(45^{\circ} - θ)$ , the horizontal range is :
$\begin{array}{l} R_{1} = \frac{u^{2} \sin [2 (45^{\circ} - θ)]}{g} = R_{1} = \frac{u^{2} \sin (90^{\circ} - 2 θ)}{g} \\ = \frac{u^{2} \cos 2 θ}{g} \end{array}$
For angle of projection $(45^{\circ} + θ)$ , the horizontal range is :
$\begin{array}{l} R_{2} = \frac{u^{2} \sin [2 (45^{\circ} + θ)]}{g} = \frac{u^{2} \sin (90^{\circ} + 2 θ)}{g} \\ = \frac{u^{2} \cos 2 θ}{g} \\ ∴ \frac{R_{1}}{R_{2}} = \frac{u^{2} \cos 2 θ / g}{u^{2} \cos 2 θ / g} = \frac{1}{1} . \end{array}$

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