76_PartUniversity Physics Solution

# 76_PartUniversity Physics Solution - 3-6 Chapter 3 (d) The...

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3-6 Chapter 3 (d) The graphs are given in Figure 3.9c Figure 3.9c EVALUATE: In the x -direction, 0 x a = and x v is constant. In the y -direction, 2 9.80 m/s y a =− and y v is downward and increasing in magnitude since y a and y v are in the same directions. The x and y motions occur independently, connected only by the time. The time it takes the book to fall 0.600 m is the time it travels horizontally. 3.10. IDENTIFY: The bomb moves in projectile motion. Treat the horizontal and vertical components of the motion separately. The vertical motion determines the time in the air. SET UP: The initial velocity of the bomb is the same as that of the helicopter. Take y + downward, so 0 x a = , 2 9.80 m/s y a =+ , 0 60.0 m/s x v = and 0 0 y v = . EXECUTE: (a) 2 1 00 2 y y yy v t a t −= + with 0 300 m yy −= gives 0 2 2( ) 2(300 m) 7.82 s 9.80 m/s y t a == = . (b) The bomb travels a horizontal distance 2 1 2 (60.0 m/s)(7.82 s) 470 m xx xx v t = = . (c) 0 60.0 m/s vv . 2 0 (9.80 m/s )(7.82 s) 76.6 m/s y vv a t =+= = . (d) The graphs are given in Figure 3.10. (e) Because the airplane and the bomb always have the same x -component of velocity and position, the plane will be 300 m directly above the bomb at impact. EVALUATE: The initial horizontal velocity of the bomb doesn&t affect its vertical motion. Figure 3.10 3.11. IDENTIFY: Each object moves in projectile motion. SET UP: Take y + to be downward. For each cricket, 0 x a = and 2 9.80 m/s y a . For Chirpy, 0 xy . For Milada, 0 0.950 m/s x v = , 0 0 y v = EXECUTE: Milada’s horizontal component of velocity has no effect on her vertical motion. She also reaches the ground in 3.50 s. 2 1 2 (0.950 m/s)(3.50 s) 3.32 m t = = EVALUATE: The x and y components of motion are totally separate and are connected only by the fact that the time is the same for both. 3.12. IDENTIFY: The person moves in projectile motion. She must travel 1.75 m horizontally during the time she falls 9.00 m vertically. SET UP: Take y + downward. 0 x a = , 2 9.80 m/s y a . x = , 0 0 y v = . EXECUTE: Time to fall 9.00 m: 2 1 2 y y t gives 0 2 2( ) 2(9.00 m) 1.36 s 9.80 m/s y t a = . Speed needed to travel 1.75 m horizontally during this time: 2 1 2 x x x xv t gives 0 1.75 m 1.29 m/s 1.36 s x t = = . EVALUATE: If she increases her initial speed she still takes 1.36 s to reach the level of the ledge, but has traveled horizontally farther than 1.75 m. 3.13. IDENTIFY: The car moves in projectile motion. The car travels 21.3 m 1.80 m 19.5 m = downward during the time it travels 61.0 m horizontally. SET UP: Take y + to be downward. 0 x a = , 2 9.80 m/s y a . x = , 0 0 y v = .

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Motion in Two or Three Dimensions 3-7 EXECUTE: Use the vertical motion to find the time in the air: 2 1 00 2 y y yy v t a t −= + gives 0 2 2( ) 2(19.5 m) 1.995 s 9.80 m/s y yy t a == = Then 2 1 2 x x x xv t gives 0 61.0 m 30.6 m/s 1.995 s x xx vv t = = . (b) 30.6m s x v = since 0 x a = . 0 19.6m s y vv a t =+= .
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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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76_PartUniversity Physics Solution - 3-6 Chapter 3 (d) The...

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