Question

A particle of mass 2 kg is dropped from a height 80 m and after striking ground, it rebound to a height of 5m. Time of contact = 0.4 sec. Find average force exerted by ground on particle.
(Please answer if you are sure and explain concept as well)​

Answer 1

Answers

Explanation:

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Answer 2

Given info : A particle of mass 2 kg is dropped from a height 80 m and after striking ground, it rebound to a height of 5m. Time of contact = 0.4 sec.

To find : the average force exerted by ground on particle is ..

solution : velocity of particle before striking , v₁ = $\sqrt{2gH}$

here, the initial height of the particle, H = 80 m

acceleration due to gravity, g = 10 m/s²

so, v₁ = $\sqrt{2\times10\times80}=40$ m/s

now velocity of the particle after striking, v₂ = $\sqrt{2gh}$

final height of the particle, h = 5m

so, v₁ = $\sqrt{2\times10\times5}=10$ m/s

change in linear momentum = impulse

⇒ final linear momentum - initial linear momentum = F × t

⇒ m(v₂ - v₁) = F × t

⇒ 2 kg(10m/s - 40m/s) = F × 0.4 s

⇒ - 150 kg m/s² = -150 N = F

now the force exerted by the ground = - force exerted by the particle = -(-150) = 150N

therefore the force exerted by ground on the particle is 150 N.