66_PartUniversity Physics Solution

# 66_PartUniversity Physics Solution - 2-36 Chapter 2 165 m/s...

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2-36 Chapter 2 EVALUATE: 165 m/s 369 mi/h = , so only Superman could jump downward with this initial speed. Figure 2.90 2.91. IDENTIFY: Apply constant acceleration equations to the motion of the rocket and to the motion of the canister after it is released. Find the time it takes the canister to reach the ground after it is released and find the height of the rocket after this time has elapsed. The canister travels up to its maximum height and then returns to the ground. SET UP: Let y + be upward. At the instant that the canister is released, it has the same velocity as the rocket. After it is released, the canister has 2 9.80 m/s y a =− . At its maximum height the canister has 0 y v = . EXECUTE: (a) Find the speed of the rocket when the canister is released: 0 0 y v = , 2 3.30 m/s y a = , 0 235 m yy −= . 22 00 2( ) y vv a =+ gives 2 0 2 ( ) 2(3.30 m/s )(235 m) 39.4 m/s va y y = = . For the motion of the canister after it is released, 0 39.4 m/s y v , 2 9.80 m/s y a , 0 235 m . 2 1 2 y y yy v t a t −= + gives 235 m (39.4 m/s) (4.90 m/s ) tt . The quadratic formula gives 12.0 s t = as the positive solution. Then for the motion of the rocket during this 12.0 s, 2 11 235 m (39.4 m/s)(12.0 s) (3.30 m/s )(12.0 s) 945 m t = + + = . (b) Find the maximum height of the canister above its release point: 0 39.4 m/s y v , 0 y v = , 2 9.80 m/s y a . ) y gives 2 0 0 2 0 (39.4 m/s) 79.2 m ( 9 . 8 0 m / s ) y vv a = = . After its release the canister travels upward 79.2 m to its maximum height and then back down 79.2 m 235 m + to the ground. The total distance it travels is 393 m. EVALUATE: The speed of the rocket at the instant that the canister returns to the launch pad is 2 0 39.4 m/s (3.30 m/s )(12.0 s) 79.0 m/s y t =+= + = . We can calculate its height at this instant by ) y with 0 0 y v = and 79.0 m/s y v = . 2 0 0 2 (79.0 m/s) 946 m ( 3 . 3 0 m / s ) y a = = , which agrees with our previous calculation. 2.92. IDENTIFY: Both objects are in free-fall and move with constant acceleration 2 9.80 m/s , downward. The two balls collide when they are at the same height at the same time. SET UP: Let y + be upward, so 2 9.80 m/s y a for each ball. Let 0 y = at the ground. Let ball A be the one thrown straight up and ball B be the one dropped from rest at height H . 0 0 A y = , 0 B yH = . EXECUTE: (a) 2 1 2 y y t applied to each ball gives 2 1 0 2 A yv tg t and 2 1 2 B yHg t . AB = gives 0 vt g t H g t and 0 H t v = . (b) For ball A at its highest point, 0 yA v = and 0 y t gives 0 v t g = . Setting this equal to the time in part (a) gives 0 0 H v vg = and 2 0 v H g = . EVALUATE: In part (a), using 0 H t v = in the expressions for A y and B y gives 2 0 1 2 gH yyH v ⎛⎞ == − ⎜⎟ ⎝⎠ . H must be less than 2 0 2 v g in order for the balls to collide before ball A returns to the ground. This is because it takes ball A

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Motion Along a Straight Line 2-37 time 0 2 v t g = to return to the ground and ball B falls a distance 2 2 0 1 2 2 v gt g = during this time. When 2 0 2 v H g = the two balls collide just as ball A reaches the ground and for H greater than this ball A reaches the ground before they collide.
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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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66_PartUniversity Physics Solution - 2-36 Chapter 2 165 m/s...

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