Discussion on Kinematics


 

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Posted on: 04:01:24 PM on September 8, 2016
Posted By:
Shweta Mishra
a ball is dropped onto the floor of 4.00 m. It rebounds to a ht. of 3.00m. If the was in contact with the floor for .010sec, what was it's avg. acceleration during contact?


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Posted on: 09:47:03 PM on September 8, 2016
Velocity of ball before hitting the ground = u = -√2gH So, u = -4√5 (taking g=10 m∕s^2 ) Height of rebound = h = ⅇ^2 H Given H= 4m and h = 3m Therefore, ⅇ^2=3/4 or ⅇ=√3/2 Now, we know e = velocity of separation (v) / velocity of approach (u) putting values , √3/2=ν/(4√5) v = 2√15 Contact with ground was for 0.01 sec So, avg acceleration = [ v - u ] / 0.01 Or , acceleration = (2√15−(−4√5))/0.01



Posted on: 05:30:56 PM on February 15, 2017
it comes about 1664 ms^-2



Posted on: 03:20:13 AM on February 18, 2017
Approx 1660





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