First slide

Projection Under uniform Acceleration

Question

A body is projected up a smooth inclined plane with velocity V from the point A 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? Length of inclined plane is 20 $\sqrt{2}$ m.

Moderate

A

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

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

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

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

Solution

Angle of projection from B is $45^{o}$ . As the body is able to cross the well of diameter 40 m, hence
$or R = \frac{v^{2}}{g} or v = \sqrt{gR}$
v = $\sqrt{10 \times 40} = 20 {ms}^{- 1}$
On the inclined plane, the retardation is:
$g \sin α = g \sin 45^{0} = \frac{10}{\sqrt{2}} {ms}^{- 2}$
Using $v^{2} - u^{2} = 2 ax$
${(20)}^{2} - u^{2} = 2 \times (- \frac{10}{\sqrt{2}}) \times 20 \sqrt{2}$
u = 20 $\sqrt{2} {ms}^{- 1} i . e ., V = 20 \sqrt{2} {ms}^{- 1}$

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