First slide

Projectile motion

Question

In figure, the angle of inclination of the inclined plane is 30°. Find the horizontal velocity V₀ so that the particle hits the inclined plane perpendicularly.

Moderate

A

$V_{0} = \sqrt{\frac{2 gH}{5}}$
B

$V_{0} = \sqrt{\frac{2 gH}{7}}$
C

$V_{0} = \sqrt{\frac{gH}{5}}$
D

$V_{0} = \sqrt{\frac{gH}{7}}$

Solution

In a direction along the inclined plane,
$0 = V_{0} \cos 30^{\circ} - gsin 30^{\circ} t \Rightarrow t = \sqrt{3} V_{0} / g$
In a direction perpendicular to incline,
$- Hcos 30^{\circ} = - Vsin 30^{\circ} t - \frac{1}{2} gcos 30^{\circ} t^{2}$
Putting the value of t and solving, we get $V_{0} = \sqrt{\frac{2 gH}{5}} .$

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