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A boy is whirling around a stone tied at the end of a 0.55 m long string at constant angular speed of 5 revolutions per second in a plane perpendicular to the ground. The string breaks when it is making an angle of 53 degrees above the horizontal and the stone is on the way up. The stone is 1.15 m above the ground when this happens. How long does the stone take to hit the ground after the string breaks?

i tried and I got 2.8972 seconds but it was wrong =(
Thank you!


    When the stone takes off, it starts out at at angle of 37 degrees above the horizontal. (That is the complement of 53 degrees). The speed is initially
    2*pi*0.55*5 = 17.28 m/s, and the vertical speed component is initially
    17.28 sin 37 = 10.40 m/s

    The stone spends 10.40/g = 1.06 s going up and an equal time going down, to the elevation where it started, for a total of 2.12 s in the air. You will need to add additional time to that to account for traveling from 0.55 + 0.55 sin 53 = 0.989 meters (the release altitude) to the ground. That is about 0.09 seconds more, for a total of 2.21 seconds


    Thank you so much!!!
    I took the sin of 53 deg instead of 37 deg

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