On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25m/s at an angle of 35º above the horizontal. If she is in flight for 1.20s, how high above the water was she when she let go of the rope?

Question

Carissa Beard

Physics

15 July, 08:41

On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25m/s at an angle of 35º above the horizontal. If she is in flight for 1.20s, how high above the water was she when she let go of the rope?

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Answers (1)

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Sabrina Graves · Answer 1

5.52 m

Explanation:

Using the equation;

Vy = Vo sin θ

Where; Vo = 2.25 m/s and θ = 35°

We get;

= 2.25 sin 35°

= 1.29 m/s

Thus; Vy = 1.29 m/s

But a = - 9.81 m/s (against gravity)

t = 1.20 s and

y = Vyt + 1/2 at²

where y is the height above the water

= 1.29 * 1.2 + (1/2 * - 9.81 * 1.2²)

= 1.548 + (-7.0632)

= - 5.5152 m

Thus; the height of the girl above the water is 5.52 m