Question

A projectile is fired with an initial speed of $5\text{m/s}$ at an angle of ${45}^{\circ }$with the horizontal, up an inclined plane having angle of inclination ${30}^{\circ }$ from horizontal. What is the range of this projectile?
(Take $g=10m/s^2$)

• A. $0.5\text{m}$
• B. $1.05\text{m}$
• C. $2.05\text{m}$
• D. $2.5\text{m}$

Answer 1

The correct option is B $1.05\text{m}$

$\alpha =$ angle of projection from horizontal$={45}^{\circ }$
$\beta =$angle of the inclined plane from horizontal$={30}^{\circ }$
The range of a projectile on an inclined plane is given by:

$R=\frac{2u^2\sin (\alpha -\beta )\cos \alpha }{g\cos \beta }\Rightarrow R=\frac{2\times 5^2\times \sin (45-30)\times \cos 45}{g\times \cos 30}$

Applying formula of $\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\Rightarrow R=\frac{50\times (\sin 45\times \cos 30-\cos 45\times \sin 30)\times \frac{1}{\sqrt{2}}}{10\times \frac{\sqrt{3}}{2}}\Rightarrow R=1.05m$