Question

a girl standing on a bridge throws a stone vertically upwards at 6 ms⁻1. it hits the water below the bridge after 2 seconds. find the speed at which the stone hits the water and the initial height of the stone

Answer 1

The speed at which the stone hits the water is 13.6m/s and the initial height of the stone is 7.6m.

Speed of stone as it hits the water

Initial velocity of stone =  6 ms⁻¹

At the maximum height, the final velocity, v = 0

Acceleration of the stone, g = 9.8 ms⁻²

Time taken to reach maximum height = t₁

• Using, v = u + gt₁

t₁ = v - u/g

since the stone is travelling upwards, g = -9.8 m/s⁻²

t₁ = 0 - 6/-9.8

t₁ = 0.61 s

Time take to fall from maximum height, t = 2 - 0.61s

Time take to fall from maximum height = 1.39s

• Calculating the final velocity using the formula, v = u + gt

where u = 0 m/s

v = 0 + 9.8 * 1.39

v = 13.6 m/s

Initial height of stone

Initial height of stone = Height of bridge, H

Time taken by the stone to fall down from height, H in water, t = 2s

Initial velocity of stone = 6 ms⁻¹

Acceleration of the stone, g = 9.8 ms⁻²

• Height of the bridge, H = -ut + gt²/2

• Initial velocity is negative since it is against gravity

H =  -6 * 2 + 9.8 * 2² /2

H = 7.6m

Therefore, the speed at which the stone hits the water is 13.6m/s and the initial height of the stone is 7.6m

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