Question

A projectile is fired horizontally from an inclined plane (of inclination ${30}^{\circ }$ with the horizontal) with speed $50\text{m/s}$. If $g=10{\text{m/s}}^2$, the range measured along the incline is

• A. $500\text{m}$
• B. $\frac{1000}{3}\text{m}$
• C. $200\sqrt{2}\text{m}$
• D. $100\text{m}$

Answer 1

The correct option is B $\frac{1000}{3}\text{m}$


Taking the axes as shown in figure,
$\overrightarrow{a}=g\sin {30}^{\circ }\hat{i}-g\cos {30}^{\circ }\hat{j}$,
$\overrightarrow{u}=50\cos {30}^{\circ }\hat{i}+50\sin {30}^{\circ }\hat{j}$

Using equations of motion
$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t;\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$
$\therefore \overrightarrow{s}=(50\cos {30}^{\circ }t+\frac{1}{2}g\sin {30}^{\circ }{t}^2)\hat{i}$
$+(50\sin {30}^{\circ }t-\frac{1}{2}g\cos {30}^{\circ }{t}^2)\hat{j}$

$x=R$ at $t=T$ (time of flight on incline)
$T=\frac{2u_y}{g_y}=\frac{2\times 50\sin {30}^{\circ }}{g\cos {30}^{\circ }}$
$\Rightarrow T=\frac{2\times 50\sin {30}^{\circ }}{10\cos {30}^{\circ }}=\frac{5\times 2}{\sqrt{3}}=\frac{10}{\sqrt{3}}$
Therefore,
$R=25\sqrt{3}\times \frac{10}{\sqrt{3}}+\frac{1}{2}\times 10\times \frac{1}{2}\times \frac{100}{3}=\frac{1000}{3}$