Question

Just before striking the ground a 0.5 kg body has a kinetic energy of 980 J. If friction is ignored from what height it was dropped.

Answer 1

Just before striking the ground a 0.5 kg body has a kinetic energy of 980 J. If friction is ignored from what height it was dropped....

answer : 200m

explanation : it is given that,

mass of body , m = 0.5 kg

kinetic energy , K.E = 980 J

from conservation of energy,

kinetic energy of body before striking the ground = potential energy of body at height h from the ground.

or, 980J = mgh

or, 980 J = 0.5 kg × 9.8 m/s² × h

or, 100/0.5 = h

or, h = 200 m

hence, height of body was 200m

Answer 2

Answer:

200 m

Explanation:

Since , mass of the body is given as 0.5 kg

The kinetic energy is given as 980 J.

We have to find the height from which it was being dropped.

Let the height be h.

From the law of conservation of energy;

kinetic energy of body before striking the ground = potential energy of body at height h from the ground.

∴ Equating both the energy;

$980 = mass \times g \times h$

$980 = 0.5 \times 9.8 \times h$

$h = \frac{980}{0.5 \times 9.8}$

∴ h = 200 m

∴ The height from which the object was dropped was 200 m