Question

A body thrown vertically upwards from the ground reaches a maximum height ‘h’. The ratio of the kinetic and potential energies of the body at a height 40\% of h from the ground is

• A. 2:3
• B. 3:2
• C. 1:1
• D. 4:9 \bigskip

Hint:

At half the maximum height, the vertical velocity decreases due to gravity, but the horizontal velocity remains unchanged. Using kinematic equations and velocity components, the total speed can be determined.

Answer 1

The Correct Option is B

When a body is thrown vertically upwards, its total energy at any point is the sum of its kinetic and potential energies. The total energy at the maximum height is equal to the potential energy at that height. The total energy of the body is conserved throughout its motion. At a height \( h' = 0.4h \), the potential energy is given by: \[ U = mgh' = mg(0.4h), \] and the kinetic energy is: \[ K = \frac{1}{2}mv^2, \] where \( v \) is the velocity at height \( h' \), and the total energy is constant at all points. Since \( v^2 = u^2 - 2gh' \) (where \( u \) is the initial velocity and \( h' \) is the height), we can substitute the value of \( h' \) and find the kinetic and potential energies. After solving, the ratio of the kinetic and potential energies at height 40\% of \( h \) is \( 3:2 \).