Unformatted text preview: 47 (a) At the highest point the velocity of the ball is instantaneously zero. Take the y axis to be 2 2 upward, set v = 0 in v 2 = v0  2gy, and solve for v0 : v0 = 2gy. Substitute g = 9.8 m/s and y = 50 m to get v0 = 2(9.8 m/s )(50 m) = 31 m/s .
2 (b) It will be in the air until y = 0 again. Solve y = v0 t  1 gt2 for t. Since y = 0 the two 2 solutions are t = 0 and t = 2v0 /g. Reject the first and accept the second: t= (c) y 60 (m) 40 20 0
...... ............ ...... ........ ..... ... ... ... ... ... ... .. ... .. .. .. .. .. . .. .. .. .. .. .. .. .. .. .. .. .. .. . .. .. .. . .. .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2v0 2(31 m/s) = 6.4 s . = 2 g 9.8 m/s v 40 ... ... ... ... ... ... (m/s) ... ... ... ... ... ... ... ... 20 ... ... ... .. 0 8 t (s) 20 40 2 4 6 t (s) 8 2 4 6 ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. 2 4 6 8 t (s) The acceleration is constant while the ball is in 2 flight: a = 9.8 m/s . Its graph is as shown on the right. 0 a 5 (m/s2 ) 10 15 ...
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This note was uploaded on 03/20/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas.
 Spring '07
 KOPP

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