Question

The maximum range of a projectile is $500$ m. If the particle is thrown up a plane is inclined at an angle of $30^{o}$ with the same speed, the distance covered by it along the inclined plane will be:

• A. 250 m
• B. 500 m
• C. 750 m
• D. 100 m

Answer 1

Answer: (B)

For the maximum range, $\theta=45^{\circ}$
$R=\frac{u^{2} sin \,2\theta}{g}=\frac{u^{2}}{g}$ sin $90^{\circ}=\frac{u^{2}}{g}$
or $500=\frac{u^{2}}{g}$
The distance covered along the inclined plane can be obtained using the equation
$v^{2} -u^{2} =2as$
or $0-u^{2} =2(-g\,\,sin\,\,30^{\circ})s$
or $s=\frac{u^{2}}{g} =500\, m$