Question

From a point on the ground a ball is projected vertically up with speed ‘v’. It reaches a point P on its path where its kinetic and potential energies are in the ratio 9: 16. The speed of the particle at P must be​

Answer 1

Given:

Initial speed of the ball when projected = v

Ratio of kinetic energy and potential energy at a point P on it path = 9:16

To find:

The speed of the particle at point P.

Solution:

As we know that kinetic energy of the ball will be 1/2*m*v²

And the potential energy of the ball at any point will be m*g*h

where m = mass of the ball

v = velocity of the ball

h = height of the ball from the ground

g = acceleration due to gravity

Since it is given that the ratio of kinetic energies and potential energies is 9/16

(1/2mv²)/ mgh = 9/16

mv²/2gh = 9/16

v = (18 gh/ 16m)^1/2

Putting the value of g = 10m/s²

v = (11.25 h/m)^1/2

Therefore the velocity of the ball at point P which is at a height h from the ground and has mass m will be (11.25 h/m)^1/2.