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18 January, 12:31

A ball is released at the top of a ramp at t = 0. which is the speed of the ball at t=4

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  1. 18 January, 12:45
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    Accel; eration due to gravity = 9.8m/s^2. Initial velocity = 0 (given). Time=4. Applying kinematic equation v=u+at, v = 0 + 9.8x4 = 39.2. Therefore speed of the ball after t=4 is 39.2
  2. 18 January, 12:58
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    AbhiGhost your answer would be right if we were talking about free fall. The force that is providing the acceleration is Fx=mgsinФ, where Ф is the angle the ramp makes with the horizontal.

    Fx=ma=mgsinФ

    Hence:

    a=gsinФ.

    And the speed is v=a*t=gsinФ*t.
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