# A ball is thrown downward from a height of 30 m with a velocity of 10 m/s. Determine the velocity with which the ball strikes the ground by using law of conservation of energy.

2
by Ishi262

2015-09-24T22:37:56+05:30
Initial height, h₁ = 30m
initial velocity, v₁ = 10 m/s
Initial total energy = PE₁ + KE₁ = mgh₁ + 1/2 mv₁²
⇒ E₁ = m×10 × 30 + 1/2 m× 10²
⇒ E₁ = 300m + 50 m
⇒ E₁ = 350 m

When it reaches ground,
Final height, h₂ = 0m
Final velocity, v₂ = v m/s
Final total energy = PE₂ + KE₂ = mgh₂ + 1/2 mv₂²
⇒ E₂ = m×10 × 0 + 1/2 m× v²
⇒ E₂ = 0 + 1/2 mv²
⇒ E₂ = 1/2 mv²

According to conservation of energy,
Initial total enegy = final total enery
⇒ E₁ = E₂
⇒ 350 m = 1/2 mv²
⇒ 350 = 1/2 v²
⇒ v² = 350*2 = 700
⇒ v = √700 m/s
⇒ v = 10√7 m/s

So the velocity with which the ball strikes the ground is 10√7 m/s.
2015-09-24T23:04:38+05:30
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