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Posted by on Tuesday, January 17, 2012 at 1:05am.

A ball is tossed so that it bounces off the ground, rises to a height of 0.90 m, and then hits the ground again 0.50 m away from the first bounce.
How long is the ball in the air between the two bounces?


What is the ball's velocity in the x-direction?


What is the ball's speed just before the second bounce?


What is the angle of the velocity vector with respect to the ground right after the first bounce?

  • physics - , Tuesday, January 17, 2012 at 3:07pm

    s = 1/2 at^2
    time to rise .9m:
    .9 = 1/2 (9.8) t^2
    t = .428 sec

    .5m/.428s = 1.17m/s

    v = at = 9.8*.428 = 4.19m/s

    h = -14.4x^2 + 7.2x
    h' = -28.8x + 7.2
    h'(0) = 7.2
    arctan(7.2) = 82 degrees

  • physics - , Tuesday, January 17, 2012 at 3:16pm

    what is acceleration ?

  • internat├Čonal school of lusaka - , Tuesday, January 17, 2012 at 3:23pm

    what is distance?

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