Question

The range of a particle projected at an angle of 15? with the horizontal is 1.5 km. Its range when projected with the same velocity at an angle of 45? with the horizontal is

• A. 1.5 km
• B. 3 km
• C. 0.75 km
• D. 4.5 km

Answer 1

Answer: (B)

3 km

Answer 2

We know, $R = \frac{u^2 \, \sin \, 2\theta}{g}$
Here, $\theta = 15^\circ, R = 1.5 \, km = 1500 \, m$
$\therefore \:\:\: 1500 = \frac{u^2 }{g} \sin 30^\circ$ or , $\frac{u^2}{g} = 3000 \, m$
Now for same velocity and $\theta = 45?$, range of the projectile would be,
$R' = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g} = 3000 \, m = 3 \, km$