Question

A body is projected up a smooth inclined plane (length = $20\sqrt{2}m$ ) with velocity u from the point M as shown in the figure. The angle of inclination is ${45}^{\circ }$ and the top is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v

• A. $40m{s}^{-1}$
• B. $40\sqrt{2}m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $20\sqrt{2}m{s}^{-1}$

Answer 1

The correct option is D

$20\sqrt{2}m{s}^{-1}$

At point N angle of projection of the body will be ${45}^{\circ }$. Let velocity of projection at this point is v. If the body just manages to cross the well then

Range = Diameter of well

$\frac{v^{2}sin2\theta }{g}=40[As\theta ={45}^{\circ }]v^2=400\Rightarrow v=20m/s$

But we have to calculate the velocity (u) of the body at point M.

For motion along the inclined plane (from M to N)

Final velocity (v) = 20 m/s, acceleration (a) = $-gsin\alpha =-gsin{45}^{\circ }$, distance of inclined plane (s) = $20\sqrt{2}m$

$(20{)}^2=u^2-2\frac{g}{\sqrt{2}}.20\sqrt{2}[Using{v}^2=u^2+2as]u^2={20}^2+400\Rightarrow u=20\sqrt{2}m/s.$