Question

The range of a particle when thrown at angle of ${15}^{\circ }$ with the horizontal is 1.5 km. What maximum range can the particle achieve if speed of projection is kept constant?

• A. 1.5 km
• B. 3.0 km
• C. 6 km
• D. 0.75 km

Answer 1

The correct option is B

3.0 km

when thrown at ${15}^{\circ }$
R = $\frac{u^{2}sin2\theta }{g}$
R = $\frac{u^{2}sin{30}^{\circ }}{g}$ ; R = $\frac{u^2}{2g}$
1500m = $\frac{u^2}{2g}$ ------(1)
When thrown at ${45}^{\circ }$, (for maximum range, as sin ${90}^{\circ }$=1 is the maximum value possible)
R = $\frac{u^{2}sin{90}^{\circ }}{g}$
R = $\frac{u^2}{g}$
From (1)
R = 2 $\times$1500m
R = 3000m = 3.0 km