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)

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