First slide

Projectile motion

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?

Difficult

A

$40 {ms}^{- 1}$
B

$40 \sqrt{2} {ms}^{- 1}$
C

$20 {ms}^{- 1}$
D

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

Solution

At point N, the angle of projection of the body will be 4 $5^{0}$ .Let,the velocity of projection at this point is v. If the body just manages to cross the well, then
Range = Diameter of well
$\begin{array}{l} \frac{v^{2} \sin 2 θ}{g} = 40 \\ [As θ = 45^{\circ}] \\ v^{2} = 400 \\ \Rightarrow v = 20 {ms}^{- 1} \end{array}$
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 ms,
acceleration $(a) = - g \sin α = - g \sin 45^{\circ}$ , distance of inclined plane $(s) = 20 \sqrt{2} m$
$\begin{array}{l} (20) = u^{2} - 2 \frac{g}{\sqrt{2}} \cdot 20 \sqrt{2} [Using v^{2} = u^{2} + 2 a s] \\ u^{2} = 20^{2} + 400 \Rightarrow 20 \sqrt{2} m / s \end{array}$

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