A particle is projected from bottom of inclined plane with speed u find distance covered along plane before coming to rest :
Show Hint
For problems on inclined planes, it is almost always easiest to resolve forces and motion into components parallel and perpendicular to the plane.
The acceleration of any object sliding on a smooth inclined plane (due to gravity alone) is always \(g \sin\theta\) directed down the plane. Memorizing this can save time.
To solve this problem, we need to determine the distance covered by a particle projected from the bottom of an inclined plane with an initial speed \( u \) before it comes to rest. The plane is inclined at an angle \( \theta \) with the horizontal.
The only force acting against the motion along the inclined plane is the component of gravity. This component is \( g \sin \theta \), where \( g \) is the acceleration due to gravity.
Using the equations of motion, the final velocity \( v \) at the point where the particle comes to rest is 0. The initial velocity \( u \) and the acceleration is \( -g \sin \theta \) (negative because it opposes the motion).
The equation of motion is:
\(v^2 = u^2 + 2as\)
Substituting the known values (with \( v = 0 \), \( a = -g \sin \theta \)), we get:
\(0 = u^2 - 2g \sin \theta \cdot s\)
Solving for \( s \):
\(u^2 = 2g \sin \theta \cdot s\)
Therefore, the distance \( s \) is:
\(s = \frac{u^2}{2g \sin \theta}\)
Thus, the correct answer is \(\frac{u^2}{2g \sin \theta}\), which matches the first option.
