A smooth plane AB is 10 meters long, It is inclined at 30 degrees to the horizontal with the lower end, B, 6 meteres vertically above horixontal ground. A particle is placed on the plane at the upper end, A, and then released from erst so that it slides down the plane. Find the speed of the particle as it strikes the ground.

Consider energy.

Gravitation energy in ball=KE when it hits

mg6+mg(10sin30)=1/2 m v^2
solve for v

To find the speed of the particle as it strikes the ground, we need to consider the conservation of energy. The initial potential energy at point A will convert into kinetic energy as the particle slides down the plane. We can calculate the speed by equating the initial potential energy to the final kinetic energy.

First, let's calculate the initial potential energy (PE) at point A.
PE = m * g * h

Where:
m = mass of the particle (assumed to be constant)
g = acceleration due to gravity (9.8 m/s^2)
h = height of point A above the ground = 6 meters

Now, let's calculate the potential energy at point A:
PE = m * g * h
PE = m * 9.8 * 6
PE = 58.8m

Next, we need to calculate the final kinetic energy (KE) of the particle just before it strikes the ground. The kinetic energy is given by:

KE = 0.5 * m * v^2

Where:
v = speed of the particle

Now, equating the initial potential energy to the final kinetic energy:
PE = KE

58.8m = 0.5 * m * v^2

Simplifying the equation:
2 * 58.8 = v^2
v^2 = 117.6

Taking the square root of both sides:
v = √(117.6)

Calculating the value of v:
v ≈ 10.85 m/s

Therefore, the speed of the particle as it strikes the ground is approximately 10.85 m/s.