Question

A body is projected up a smooth inclined plane with a velocity $v_{0}$ from point $A$ as shown in the figure. The angle of inclination is $45^\circ $ and top $B$ of the plane is connected to a well of diameter $40 \, m$ . If the body just manages to cross the well, the value of $v_{0}$ is (Length of the inclined plane is $20\sqrt{2} \, m$ , and $g=10 \, m \, s^{- 2}$ )

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

Answer 1

Answer: (B)

Let $v_{2}$ be the velocity at point B at it cross the well. thus the range of the projection motion.
$\textit{R}=\frac{u^{2} \text{sin 2} \text{\theta }}{g}=\frac{v_{2}^{2} sin 2 \text{\theta }}{g}=\text{40 m}$
also as the inclined plane is at an angle $45^{\text{ο}}$ thus $\theta $ for projection motion is also $45^{\text{ο}}$
thus $\frac{v_{2}^{2} sin 2 \text{\theta }}{g} = \text{40} = \frac{v_{2}^{2} sin 90^{\text{ο}}}{g} = \frac{v^{2}}{g}$
$v_{2}^{2} = \textit{40 g}$
$v_{2}^{2} = \text{400}$
$v_{2} = \text{20}$
now given initial velocity is $v_{0}$ so apply the equation of motion
$v^{2} = u^{2} + \textit{2 as}$

$\left(\text{20}\right)^{2} = v_{0}^{2} + \text{2} \left(g sin \text{\theta }\right) . \left(\text{20} \sqrt{2}\right)$
$v_{0}^{2} = \text{400} + \text{2.20. 10}$
$v_{0}=\text{20}\sqrt{2} \, ms^{- 1}$