a body is lights down if exit Garba track that is 1 quadrant of circle of radius R if there is no friction on the body starts from rest its speed at the bottom of the track is

this makes no sense to me.

Me neither

To find the speed of the body at the bottom of the track, we can use the concept of conservation of energy.

When the body is at the topmost point of the track (quadrant of the circle) and ready to move, its initial potential energy is converted to kinetic energy at the bottom of the track.

The potential energy of an object at a certain height is given by the formula:

Potential Energy (PE) = mass (m) * gravity (g) * height (h)

In this case, the height (h) is equal to the radius (R) of the circle.

The kinetic energy of an object is given by:

Kinetic Energy (KE) = (1/2) * mass (m) * velocity^2

Since the body starts from rest, its initial velocity (v_initial) is 0.

Now, using conservation of energy, we can equate the initial potential energy to the final kinetic energy:

PE = KE

m * g * R = (1/2) * m * v_final^2

Since the mass (m) cancels out, we can simplify the equation to:

g * R = (1/2) * v_final^2

Now, to find the speed at the bottom of the track (v_final), we can rearrange the equation:

v_final = sqrt(2 * g * R)

Therefore, the speed at the bottom of the track is given by the formula:

v_final = sqrt(2 * g * R)

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and R is the radius of the circle.