Question

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?

Answer 1

5.52 m

Step-by-step 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