Question

A projectile is thrown from the base of an incline of ${30}^{\circ }$ as shown in the figure. It is thrown at an angle of ${60}^{\circ }$ from the horizontal direction at a speed of $10\text{m/s}$. The total time of flight is: $(g=10{\text{m/s}}^2)$

• A. $\frac{2}{\sqrt{3}}\text{s}$
• B. $\frac{\sqrt{3}}{2}\text{s}$
• C. $\sqrt{3}\text{s}$
• D. $0.5\text{s}$

Answer 1

The correct option is A $\frac{2}{\sqrt{3}}\text{s}$

The relation of time of flight for a projectile on an inclined plane is given by,
$T=\frac{2u\sin (\theta -\alpha )}{g\cos \alpha }$
Now, $\theta ={60}^{\circ }$, $\alpha ={30}^{\circ }$, $u=10\text{m/s}$. Putting these values, we get
$T=\frac{2\times 10\sin ({60}^{\circ }-{30}^{\circ })}{g\cos {30}^{\circ }}$
$\Rightarrow T=\frac{20\sin {30}^{\circ }}{10\times \cos {30}^{\circ }}=\frac{2}{\sqrt{3}}\text{s}$