Question

The range of a projectile when fired at 75° with the horizontal is 0.5km what will be its range when it is fired at an angle 45°

Answer 1

Answer:

The range of the projectile when it is fired at 45° is 1 km

Explanation:

We know that range of projectile with initial velocity u and angle of projection α is given by

$R=\frac{u^2sin2\alpha}{g}$

When α = 75°, the Range R = 0.5 km =500 m

∴ $R=\frac{u^2sin(2\times 75^\circ)}{g}$

$\implies R=\frac{u^2 \sin 150^\circ}{g}$

$\implies R=\frac{u^2 \sin (180^\circ-30^\circ)}{g}$

$\implies R=\frac{u^2 \sin 30^\circ}{g}$                (∵ sin(180°-θ) = sinθ)

$\implies R=\frac{u^2}{2g}$                                    (∵ sin30° = 1/2)

Similarly when the angle of projection is 45°, the range R' will be

$R'=\frac{u^2sin(2\times 45^\circ)}{g}$

$\implies R'=\frac{u^2sin90^\circ}{g}$

$R'=\frac{u^2}{g}$                                                       (∵ sin90° = 1)

Now,

$\frac{R'}{R} =\frac{u^2/g}{u^2/2g}$

$\implies \frac{R'}{R} =\frac{u^2}{g} \times \frac{2g}{u^2}$

$\implies \frac{R'}{R} =2$

$\implies R' =2R$

$\implies R' =2 \times 500 \text{ m}$

$\implies R' =1000 \text{ m}$

Therefore, the range of the projectile when it is fired at 45° is 1 km