Question
On a hot summer day, a young girl swings on a rope above the local swimming hole
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 35.0 above the horizontal. If she is in flight for 0.616sec, how high above the water was she when she let go of the rope?
Answers
Her vertical velocity component at release was 2.25*sin 35 = 1.29 m/s
Solve the following equation for altitude vs. time to get the initial height H:
y = H + 1.29 t - (g/2) t^2
When t = 0.616 s, y = 0
Use that to solve for H
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