# ch02-p055 - 55. (a) We first find the velocity of the ball...

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where Δ v is the change in its velocity during contact with the ground and 3 20.0 10 s t Δ= × is the duration of contact. Now, to find the velocity just before contact, we put the origin at the point where the ball is dropped (and take + y upward) and take t = 0 to be when it is dropped. The ball strikes the ground at y = –15.0 m. Its velocity there is found from Eq. 2-16: v 2 = –2 gy . Therefore, 2 2 2(9.8 m/s )( 15.0 m) 17.1 m/s vg y =− − =− where the negative sign is chosen since the ball is traveling downward at the moment of contact. Consequently, the average acceleration during contact with the ground is
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## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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