81_PartUniversity Physics Solution

# 81_PartUniversity - Motion in Two or Three Dimensions 3-11 3.23(c v0 = 4.0 m/s gives y = 2.29 m In this case the dart was fired with so slow a

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Motion in Two or Three Dimensions 3-11 (c) 0 4.0 m/s v = gives 2.29 m y =− . In this case, the dart was fired with so slow a speed that it hit the ground before traveling the 3-meter horizontal distance. EVALUATE: For (a) and (d) the trajectory of the dart has the shape shown in Figure 3.26 in the textbook. For (c) the dart moves in a parabola and returns to the ground before it reaches the x -coordinate of the monkey. 3.23. IDENTIFY: Take the origin of coordinates at the roof and let the -direction y + be upward. The rock moves in projectile motion, with 0 x a = and . y ag = − Apply constant acceleration equations for the x and y components of the motion. SET UP: 00 0 cos 25.2 m/s x vv α == 0 sin 16.3 m/s y Figure 3.23a (a) At the maximum height 0. y v = 2 9.80 m/s , y a 0, y v = 0 16.3 m/s, y v =+ 0 ? yy = 22 2( ) y vv a EXECUTE: 2 0 0 2 0 (16.3 m/s) 13.6 m ( 9 . 8 0 m / s ) y a −= = = + (b) SET UP: Find the velocity by solving for its x and y components. 0 25.2 m/s xx (since 0 x a = ) ?, y v = 2 9.80 m/s , y a 0 15.0 m (negative because at the ground the rock is below its initial position), 0 16.3 m/s y v = ) y 2 ) y a y y + ( y v is negative because at the ground the rock is traveling downward.) EXECUTE: (16.3 m/s) 2( 9.80 m/s )( 15.0 m) 23.7 m/s y v + − Then 2 2 (25.2 m/s) ( 23.7 m/s) 34.6 m/s. xy v = + = (c) SET UP: Use the vertical motion ( y -component) to find the time the rock is in the air: t = 23.7 m/s y v (from part (b)), 2 9.80 m/s , y a 0 16.3 m/s y v EXECUTE: 0 2 23.7 m/s 16.3 m/s 4.08 s 9.80 m/s y t a −− = + SET UP: Can use this t to calculate the horizontal range: 4.08 s, t = 0 25.2 m/s, x v = 0, x a = 0 ? = EXECUTE: 2 1 2 (25.2 m/s)(4.08 s) 0 103 m xx v t a t −= + = += (d) Graphs of x versus t , y versus t , x v versus t , and y v versus t : Figure 3.23b

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3-12 Chapter 3 EVALUATE: The time it takes the rock to travel vertically to the ground is the time it has to travel horizontally. With 0 16.3 m/s y v =+ the time it takes the rock to return to the level of the roof ( 0) y = is 0 2 / 3.33 s. y tvg == The time in the air is greater than this because the rock travels an additional 15.0 m to the ground. 3.24. IDENTIFY: Consider the horizontal and vertical components of the projectile motion. The water travels 45.0 m horizontally in 3.00 s. SET UP: Let y + be upward. 0 x a = , 2 9.80 m/s y a =− . 00 0 cos x vv θ = , 000 sin y = . EXECUTE: (a) 2 1 2 x x x xv t a t −= + gives 0 (cos ) x t −= and 0 45.0 m cos 0.600 (25.0 m/s)(3.00 s) ; 0 53.1 = ° (b) At the highest point 0 (25.0 m/s)cos53.1 15.0 m/s xx ° = , 0 y v = and 22 15.0 m/s xy v = . At all points in the motion, 2 9.80 m/s a = downward. (c) Find 0 yy when 3 00s t =. : 2 11 (25.0 m/s)(sin53.1 )(3.00 s) ( 9.80 m/s )(3.00 s) 15.9 m yy v t = ° +− = 0 15.0 m/s , 2 0 (25.0 m/s)(sin53.1 ) (9.80m/s )(3.00 s) 9.41 m/s y vv a t =+= ° = , and 2 2 (15.0 m/s) ( 9.41 m/s) 17.7 m/s vvv = + = EVALUATE: The acceleration is the same at all points of the motion. It takes the water 0 2 20.0 m/s 2.04 s 9.80 m/s y y v t a = to reach its maximum height. When the water reaches the building it has passed its maximum height and its vertical component of velocity is downward. 3.25. IDENTIFY and SET UP: The stone moves in projectile motion. Its initial velocity is the same as that of the balloon.
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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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81_PartUniversity - Motion in Two or Three Dimensions 3-11 3.23(c v0 = 4.0 m/s gives y = 2.29 m In this case the dart was fired with so slow a

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