A partical p is projected from a point on the surface of long smooth inclined plane. Simultaneously another particle Q is released on the smooth inclined plane from the same position. P and Q collide after t=4 sec. The speed of projection of p is:

2 seg.

Please explain me

To find the speed of projection of particle P, we need to determine the height it is projected from on the inclined plane.

Let's assume the height from which particle P is projected is h meters from the base of the inclined plane.

The motion of particle P can be divided into two components: one along the inclined plane and the other perpendicular to it.

1. Motion along the inclined plane:
Since the inclined plane is smooth, there is no friction acting on particle P. Therefore, its motion along the inclined plane is a uniform accelerated motion (assuming the acceleration due to gravity is g m/s²).

The formula for the displacement of an object undergoing uniform accelerated motion is given by:
s = ut + 0.5at²
Here, s is the displacement, u is the initial velocity, t is time, and a is the acceleration.

Since particle P collides with particle Q after 4 seconds, the displacement along the inclined plane will be equal to h (the height it was projected from):
h = 0 + 0.5gt²
h = 0.5 × 9.8 × 4²
h = 0.5 × 9.8 × 16
h = 78.4 meters

2. Motion perpendicular to the inclined plane:
The time taken by particle P to collide with particle Q is 4 seconds. In this time, particle P would have fallen vertically under the influence of gravity. The formula for the distance fallen under constant acceleration due to gravity is given by:
s = 0.5gt²

Since the time is 4 seconds, the displacement perpendicular to the inclined plane is:
s = 0.5 × 9.8 × 4²
s = 0.5 × 9.8 × 16
s = 78.4 meters

Therefore, the height h from which particle P is projected is equal to the distance it has fallen vertically. So the speed of projection of particle P is the same as the speed it gained while falling vertical, which is equal to the final velocity after falling freely for 4 seconds.

The formula for the final velocity of an object falling under gravity is given by:
v = u + gt
Here, v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

Since the initial velocity is 0, the final velocity of particle P after falling for 4 seconds is:
v = 0 + 9.8 × 4
v = 0 + 39.2
v = 39.2 m/s

Therefore, the speed of projection of particle P is 39.2 m/s.