54. We apply Eq. 4-21, Eq. 4-22 and Eq. 4-23. (a) From Δxvtx=0, we find 040 m / 2 s20 m/s.xv==(b) FromΔyvt gty=−0122, we find ()2210253 m(9.8 m/s )(2 s)/ 236yv=+=m/s. (c) From vvgtyy′0with vy= 0 as the condition for maximum height, we obtain
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.