velocity Question Details A 0.66 kg, 0.37 m diameter ball is dropped from rest f

Question

velocity
Question Details
A 0.66 kg, 0.37 m diameter ball is dropped from rest from the origin at yo = 0.


a) What is the ball's position at time t = 4.6 s?


b) What is the ball's velocity at time t = 4.6 s?


c) How long did it take the ball to fall half the distance calculated in part (a) above?



d) What was the ball's velocity when it was at half the distance calculated in part (a) above?

e) How long did it take the ball to get to half the velocity calculated in part (b) above?

f) What was the ball's position when it had half the velocity calcualted in part (b) above?

Explanation / Answer

Hi, This is the case of a freely falling body. In all cases x = 0. only y co-ordinate will change. a) We know s = ut + (1/2)at^2.                => s = 0 + (0.5 * (9.8) * (4.6)^2)   (since g is the only acceleration acting).                => s = 103.684m. Since it started at y = 0, now it will be at y = -103.684m b) v = u + at = 0 + 9.8 * 4.6 = 45.08m/s c) half of a is 51.842m.                hence from s = ut + 1/2at^2, we have 51.842 = 0.5 * 9.8 * t^2                => t = 3.253sec d) at this point, its velocity is again v = u + at = 9.8 * 3.253 = 31.879 m/s e) half the velocity in part b = 22.54m/s.       from v = u + at we have, 22.54 = 9.8 * t                                            => t = 2.3sec f) At this instant, ball position is s = ut + (1/2)at^2 = 0.5 * 9.8 * (2.3)^2 = 25.921m.       Hence the balls position now is y = -25.921m with x as always 0. Hope this helps you.                                                                           

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