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46_PartUniversity Physics Solution

46_PartUniversity Physics Solution - 2-16 Chapter 2...

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2-16 Chapter 2 E VALUATE : We could use either 0 0 2 y y v v y y t + = or 2 2 0 0 2 ( ) y y y v v a y y = + to check our results. Figure 2.42 2.43. I DENTIFY : When the only force is gravity the acceleration is 2 9.80 m/s , downward. There are two intervals of constant acceleration and the constant acceleration equations apply during each of these intervals. S ET U P : Let y + be upward. Let 0 y = at the launch pad. The final velocity for the first phase of the motion is the initial velocity for the free-fall phase. E XECUTE : (a) Find the velocity when the engines cut off. 0 525 m y y = , 2 2.25 m/s y a = + , 0 0 y v = . 2 2 0 0 2 ( ) y y y v v a y y = + gives 2 2(2.25 m/s )(525 m) 48.6 m/s y v = = . Now consider the motion from engine cut off to maximum height: 0 525 m y = , 0 48.6 m/s y v = + , 0 y v = (at the maximum height), 2 9.80 m/s y a = − . 2 2 0 0 2 ( ) y y y v v a y y = + gives 2 2 2 0 0 2 0 (48.6 m/s) 121 m 2 2( 9.80 m/s ) y y y v v y y a = = = and 121 m 525 m 646 m y = + = . (b) Consider the motion from engine failure until just before the rocket strikes the ground: 0 525 m y y = − , 2 9.80 m/s y a = − , 0 48.6 m/s y v = + . 2 2 0 0 2 ( ) y y y v v a y y = + gives 2 2 (48.6 m/s) 2( 9.80 m/s )( 525 m) 112 m/s y v = − + = − . Then 0 y y y v v a t = + gives 0 2 112 m/s 48.6 m/s 16.4 s 9.80 m/s y y y v v t a = = = . (c) Find the time from blast-off until engine failure: 0 525 m y y = , 0 0 y v = , 2 2.25 m/s y a = + . 2 1 0 0 2 y y y y v t a t = + gives 0 2 2( ) 2(525 m) 21.6 s 2.25 m/s y y y t a = = = . The rocket strikes the launch pad 21.6 s 16.4 s 38.0 s + = after blast off. The acceleration y a is 2 2.25 m/s + from 0 t = to 21.6 s t = . It is 2 9.80 m/s from 21.6 s t = to 38.0 s . 0 y y y v v a t = + applies during each constant acceleration segment, so the graph of y v versus t is a straight line with positive slope of 2 2.25 m/s during the blast-off phase and with negative slope of 2 9.80 m/s after engine failure. During each phase 2 1 0 0 2 y y y y v t a t = + . The sign of y a determines the curvature of ( ) y t . At 38.0 s t = the rocket has returned to 0 y = . The graphs are sketched in Figure 2.43. E VALUATE : In part (b) we could have found the time from 2 1 0 0 2 y y y y v t a t = + , finding y v first allows us to avoid solving for t from a quadratic equation. Figure 2.43
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Motion Along a Straight Line 2-17 2.44. I DENTIFY : Apply constant acceleration equations to the vertical motion of the sandbag. S ET U P : Take y + upward. 2 9.80 m/s y a = − . The initial velocity of the sandbag equals the velocity of the balloon, so 0 5.00 m/s y v = + . When the balloon reaches the ground, 0 40.0 m y y = − . At its maximum height the sandbag has 0 y v = .
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