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^o 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. $40ms^{- 1}$
• B. $40\sqrt{2}ms^{- 1}$
• C. $20ms^{- 1}$
• D. $20\sqrt{2}ms^{- 1}$

Answer 1

Answer: (D)

$20\sqrt{2}ms^{- 1}$

Answer 2

At point N angle of projection of the body will be 45⁰. Let velocity of projection at this point is v.

If the body just manages to cross the well then $\text{Range} = \text{Diameterofwell}$

$\frac{v^{2}\sin 2\theta}{g} = 40$ $\left\lbrack \text{As}\theta = \text{45}{^\circ} \right\rbrack$

$v^{2} = 400$ ⇒ $v = 20m/s$

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 (1) = – g sinα = – g sin 45^o, distance of inclined plane (s) = $20\sqrt{2}$m

$(20)^{2} = u^{2} - 2\frac{g}{\sqrt{2}}.20\sqrt{2}$ [Using v² = u² + 2as]

$u^{2} = 20^{2} + 400$ ⇒ u $= 20\sqrt{2}m/s.$