Question

The range of a particle when launched at an angle of ${15}^{\circ }$ with the horizontal is $1.5\text{km}$. What is the range of the projectile when launched at an angle of ${45}^{\circ }$ to the horizontal ?

• A. $1.5\text{km}$
• B. $3.0\text{km}$
• C. $6.0\text{km}$
• D. $0.75\text{km}$

Answer 1

The correct option is B $3.0\text{km}$

Range of a projectile projected with initial velocity $u$ at an angle $\theta$ with the horizontal is given by
$R=\frac{u^2\sin 2\theta }{g}$

Range for ${15}^{\circ }$ angle of projection
$R_{{15}^{\circ }}=\frac{u^2\sin (2\times {15}^{\circ })}{g}=\frac{u^2}{2g}=1.5\text{km}$

$\therefore$ Range for ${45}^{\circ }$ angle of projection
$R_{{45}^{\circ }}=\frac{u^2\sin (2\times {45}^{\circ })}{g}=\frac{u^2}{g}=1.5\times 2=3\text{km}$