46_PartUniversity Physics Solution

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

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2-16 Chapter 2 EVALUATE: We could use either 0 0 2 yy vv t + ⎛⎞ −= ⎜⎟ ⎝⎠ or 22 00 2( ) y vv a =+ to check our results. Figure 2.42 2.43. IDENTIFY: 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. SET UP: 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. EXECUTE: (a) Find the velocity when the engines cut off. 0 525 m , 2 2.25 m/s y a , 0 0 y v = . ) 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 =− . ) y gives 2 0 0 2 0( 4 8 . 6 m / s ) 121 m ( 9 . 8 0 m / s ) 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 , 2 9.80 m/s y a , 0 48.6 m/s y v . ) y gives (48.6 m/s) 2( 9.80 m/s )( 525 m) 112 m/s y v + − . Then 0 y t gives 0 2 112 m/s 48.6 m/s 16.4 s 9.80 m/s y t a −− = . (c) Find the time from blast-off until engine failure: 0 525 m , 0 0 y v = , 2 2.25 m/s y a . 2 1 2 y y yy v t a t −= + gives 0 2 2( ) 2(525 m) 21.6 s 2.25 m/s 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 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 2 y y t . The sign of y a determines the curvature of ( ) yt . At 38.0 s t = the rocket has returned to 0 y = . The graphs are sketched in Figure 2.43. EVALUATE: In part (b) we could have found the time from 2 1 2 y y 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. IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag. SET UP: 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 yy −= . At its maximum height the sandbag has 0 y v = . EXECUTE: (a) 0.250 s t = : 22 2 11 00 (5.00 m/s)(0.250 s) ( 9.80 m/s )(0.250 s) 0.94 m yy v t a t −= + = +− = . The sandbag is 40.9 m above the ground.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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

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