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Posted by on Sunday, February 20, 2011 at 6:03am.

One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While your opponent is watching the first one, a second snowball is thrown at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 10.0 m/s. The first one is thrown at an angle of 65.0° with respect to the horizontal.
How many seconds later should the second snowball be thrown after the first to arrive at the same time?

  • physics - , Sunday, February 20, 2011 at 9:51am

    To arrive at the same PLACE, the second snowball must be launched at a complementary angle of 25 degrees.

    The difference in the time of flight for those two launch angles is the required delay time, Tdelay.

    Tdelay = 2 (Vsin65 - Vsin25)/g
    = [(10 m/s)/9.8 m/s^2]*(0.4836)
    = 0.49 seconds

  • physics - , Saturday, August 6, 2011 at 7:44am

    A particular Ferris wheel (a rigid wheel rotating in a vertical plane about a horizontal axis) at a local carnival has a radius of 20.0 m and it completes 1 revolution in 9.84 seconds.
    (a) What is the speed (m/s) of a point on the edge of the wheel?
    Using the coordinate system shown, nd:
    the (b) x component of the acceleration of point A at the top of
    the wheel.
    the (c) y component of the acceleration of point A at the top of the wheel
    the (d) x component of the acceleration of point B at the bottom of the wheel.
    the (e) y component of the acceleration of point B at the bottom of the wheel
    the (f) x component of the acceleration of point C on the edge of the wheel.
    the (g) y component of the acceleration of point C on the edge of the wheel

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