where Δvis the change in its velocity during contact with the ground and 320.0 10 st−Δ=×is the duration of contact. Now, to find the velocity just beforecontact, we put the origin at the point where the ball is dropped (and take +yupward) 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: v2= –2gy. Therefore, 222(9.8 m/s )( 15.0 m)17.1 m/svgy=− −−=−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.