First slide

Projectile motion

Question

AB is an inclined plane of inclination $30 °$ with horizontal. Point O is 20m above point A. A particle is projected horizontally and it collides perpendicular to the plane AB . Speed of the particle, at ‘O’ must be $(g = 10 {m/s}^{2})$

Moderate

A

$13 m / s$
B

$8 \sqrt{3} m / s$
C

$4 \sqrt{5} m / s$
D

$2 \sqrt{5} m / s$

Solution

Resolving initial velocity and g along x and y direction
We get $u_{x} = u \cos θ, u_{y} = u \sin {θ, θ=30}^{\circ}$
$g_{x} = - g \sin θ$
$g_{y} = g \cos θ$
From diagram
Applying equation of motion in x-direction
$v_{x} = 0$ ( $∵$ body hits the plane normally )
$u_{x} = u \cos θ$
$\Rightarrow v_{x} = u_{x} + a_{x} t$
$\Rightarrow 0 = u \cos 30 ° - g \sin 30 ° t$
$\Rightarrow u \sqrt{3} = g t$
$\Rightarrow t = \frac{\sqrt{3} u}{g}$ ……. (1)
Applying equation of motion in y direction
$\begin{array}{l} \Rightarrow x_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} \\ \Rightarrow 20 \cos 30^{\circ} = u \sin 30^{\circ} t + \frac{1}{2} g \cos 30^{\circ} t^{2} \dots \dots (2) \\ \Rightarrow 20 \sqrt{3} = u t + \frac{1}{2} g t^{2} \sqrt{3} \\ \Rightarrow 20 \sqrt{3} = \frac{u^{2} \sqrt{3}}{g} + \frac{\sqrt{3}}{2} g \cdot \frac{3 u^{2}}{g^{2}} (∵ g = 10 m / s) \\ \Rightarrow 20 \sqrt{3} = \frac{u^{2} (2 \sqrt{3} + 3 \sqrt{3})}{2 g} \\ \Rightarrow 20 \sqrt{3} = \frac{u^{2} (5 \sqrt{3})}{2 g} \\ \Rightarrow u^{2} = 80 \end{array}$
Solving the above equation we get $u = 4 \sqrt{5} m / s$
Therefore, the correct answer is (C).

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