Question

A motorcyclist starts from the bottom of a slope of angle ${45}^{\circ }$ to cross the valley $PR$ as shown in the figure. The width of the valley is $90\text{m}$ and length of the slope is $80\sqrt{2}\text{m}$. The minimum velocity at point $O$ required to clear the valley will be

• A. $70\text{m/s}$
• B. $30\text{m/s}$
• C. $50\text{m/s}$
• D. $100\text{m/s}$

Answer 1

The correct option is C $50\text{m/s}$

$R=\frac{u^2}{g}\sin 2\theta =\frac{v^2}{g}$
Velocity of take off at $P$ or $v=\sqrt{Rg}=\sqrt{90\times 10}=30\text{m/s}$
$u=\sqrt{v^2+2g\sin \theta S}$
[$u\to$ velocity at point $O$]
$u=\sqrt{(30{)}^2+2\times 10\times \frac{1}{\sqrt{2}}\times 80\sqrt{2}}=50\text{m/s}$