61_PartUniversity Physics Solution

61_PartUniversity Physics Solution - Motion Along a...

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Motion Along a Straight Line 2-31 2.76. IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. SET UP: Let y + be downward. The egg has 0 0 y v = and 2 9.80 m/s y a = . At the height of the professor&s head, the egg has 0 44.2 m yy −= . EXECUTE: 2 1 00 2 y y yy v t a t −= + gives 0 2 2( ) 2(44.2 m) 3.00 s 9.80 m/s y t a == = . The professor walks a distance (1.20 m/s)(3.00 s) 3.60 m x xx v t −= = = . Release the egg when your professor is 3.60 m from the point directly below you. EVALUATE: Just before the egg lands its speed is 2 (9.80 m/s )(3.00s) 29.4 m/s = . It is traveling much faster than the professor. 2.77. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. SET UP: Let y + be upward. At the maximum height, 0 y v = . When the rock returns to the surface, 0 0 = . EXECUTE: (a) 22 2( ) y vv a =+ gives 2 1 0 2 y y aH v =− , which is constant, so EE MM a H = . 2 E ME 2 M 9.80 m/s 2.64 3.71 m/s a H HH H a ⎛⎞ = ⎜⎟ ⎝⎠ . (b) 2 1 2 y y t with 0 0 gives 0 2 y y at v , which is constant, so aT a T = . E M 2.64 a TT T a ⎡⎤ ⎢⎥ ⎣⎦ . EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in the air for a longer time. 2.78. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. SET UP: For the thrown ball, let y + be upward. 2 9.80 m/s y a . 0 0 = when the ball returns to its original position. EXECUTE: (a) It takes her 5.50 m 2.20 s 2.50 m/s = to reach the table and an equal time to return. For the ball, 0 0 , 4.40 s t = and 2 9.80 m/s y a . 2 1 2 y y t gives 2 11 0 ( 9.80 m/s )(4.40 s) 21.6 m/s va t =− − = . (b) Find 0 when 2.20 s t = . 2 (21.6 m/s)(2.20 s) ( 9.80 m/s )(2.20 s) 23.8 m t = +− = EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling. 2.79. (a) IDENTIFY: Use constant acceleration equations, with , y ag = downward, to calculate the speed of the diver when she reaches the water. SET UP: Take the origin of coordinates to be at the platform, and take the -direction y + to be downward. 0 21.3 m, + 2 9.80 m/s , y a 0 0 y v = (since diver just steps off), ? y v = ) y EXECUTE: 2 0 2 ( ) 2(9.80 m/s )(31.3 m) 20.4 m/s. y y We know that y v is positive because the diver is traveling downward when she reaches the water. The announcer has exaggerated the speed of the diver. EVALUATE: We could also use 2 1 2 y y t to find 2.085 s. t = The diver gains 9.80 m/s of speed each second, so has 2 (9.80 m/s )(2.085 s) 20.4 m/s y v when she reaches the water, which checks.
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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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61_PartUniversity Physics Solution - Motion Along a...

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