# 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.

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by Ishi262

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by Ishi262

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The Brainliest Answer!

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 =

So the velocity with which the ball strikes the ground is 10√7 m/s.