Question

A body of mass 1 kg is thrown vertically upwards with an initial kinetic energy of 98J. The height at which Kinetic energy is reduced to one half of initial value will be
(a) 5m
(b) 10m
(c) 2.5m
(d) 1.25m​

Answer 1

Given :

Mass of body = 1 kg

Value of initial kinetic energy = 98J

To Find :

We have to find the height at which kinetic energy is reduced to one half of initial value.

Solution :

❒ Let H be the height at which kinetic energy is reduced to one half of initial value.

❒ Potential energy of body of mass m at a height of H from the ground is given by,

• U = m × g × H

Where,

» U denotes potential energy

» m denotes mass

» g denotes acc. due to gravity

» H denotes height

Such questions can be easily solved by using concept of mechanical energy conservation.

• Mechanical energy of body at ground will be equal to the mechanical energy of body at height H$.$

➠ U + K = U' + K'

ATQ, K' = K/2

• PE at ground = zero

➠ 0 + K = mgH + K/2

➠ mgH = K/2

➠ 1 × 9.8 × H = 98/2

➠ H = 10/2

➠ H = 5 m

∴ (A) is the correct answer!

Answer 2

Answer:

A body of mass 1 kg is thrown vertically upwards with an initial kinetic energy of 98J. The height at which Kinetic energy is reduced to one half of initial value will be:

〓 a.5m