On a hot summer day, a young girl swings on a rope above the local swimming hole

Question

On a hot summer day, a young girl swings on a rope above the local swimming hole (Figure 1) . When she lets go of the rope her initial velocity is 2.20{ m m/s}  at an angle of 35.0 ^circ above the horizontal.

If she is in flight for 0.617{ m s} , how high above the water was she when she let go of the rope?

I need step by step how to answer this and what formulas used, it would be helpful you showed me what information is plugged into what part of the formulas used. What information is given and what information is needed. Thank you.

Explanation / Answer

y = (Vo sin theta )t - (1/2 * g * t^2)

= (2.20 * sin 35 * 0.617 ) - (0.5 * 9.8 * 0.617^2)

= -1.09 m

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