Physics Solution Manual for 1100 and 2101

26 km 048 km a from the pythagorean theorem we have r

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Unformatted text preview: ne the magnitude and 11. REASONING direction of the bird watcher's displacement. Once the displacement is known, Equation 3.1 can be used to find the average velocity. SOLUTION The following table gives the components of the individual displacements of the bird watcher. The last entry gives the components of the bird watcher's resultant displacement. Due east and due north have been chosen as the positive directions. Displacement East/West Component North/South Component A B C 0.50 km 0 –(2.15 km) cos 35.0° = –1.76 km 0 –0.75 km (2.15 km) sin 35.0° = 1.23 km r=A+B+C –1.26 km 0.48 km a. From the Pythagorean theorem, we have r= North (–1.26 km) 2 + (0.48 km) 2 = 1.35 km r 0.48 km The angle θ is given by θ = tan −1 0.48 F km I = G km J HK 1.26 θ 1.26 km 21° , north of west b. From Equation 3.1, the average velocity is East 104 KINEMATICS IN TWO DIMENSIONS v= ∆r 1.35 km = = 0.540 km/h, 21° north of west ∆t 2.50 h Note that the direction of the average velocity is, by definition, the same as the direction of the displacement. 12. REASONING AND SOLUTION The vertical motion consists of the ball rising for a time t stopping and returning to the ground in another time t. For the upward portion t= v0 y − v y g = v0 y g Note: vy = 0 m/s since the ball stops at the top. Now V0y = v0 sin θ0 = (25.0 m/s) sin 60.0° = 21.7 m/s t = (21.7 m/s)/(9.80 m/s2) = 2.21 s The required "hang time" is 2t = 4.42 s . 13. SSM REASONING AND SOLUTION As shown in Example 3, the time required for the package to hit the ground is given by t = 2 y / a y and is independent of the plane’s horizontal velocity. Thus, the time needed for the package to hit the ground is still 14.6 s . 14. REASONING The magnitude v of the puck’s velocity is related to its x and y velocity 2 components (vx and vy ) by the Pythagorean theorem (Equation 1.7); v = v x + v 2 . The y relation vx = v0 x + axt (Equation 3.3a) may be used to find vx, since v0x, ax, and t are known. Likewise, the relation v y = v0 y + a y t (Equation 3.3b) may be employed to determine vy, since v0y, ay, and t are known. Once vx and vy are determined, the angle θ that the velocity makes with respect to the x axis can be found by using the inverse tangent ( ) function (Equation 1.6); θ = tan −1 v y / v x . Chapter 3 Problems 105 SOLUTION Using Equations 3.3a and 3.3b, we find that ( ) v y = v0 y + a y t = +2.0 m/s + ( −2.0 m/s2 ) ( 0.50 s ) = +1.0 m/s v x = v0 x + a x t = +1.0 m/s + 2.0 m/s2 ( 0.50 s ) = +2.0 m/s The magnitude v of the puck’s velocity is 2 v = vx + v2 = y ( 2.0 m/s )2 + (1.0 m/s )2 = 2.2 m/s The angle θ that the velocity makes with respect to the +x axis is vy −1 1.0 m/s = tan = 27° above the +x axis 2.0 m/s vx θ = tan −1 _____________________________________________________________________________________________ 15. REASONING Trigonometry indicates that the x and y components of the dolphin’s velocity are related to the launch angle θ according to tan θ = vy /vx. SOLUTION Using trigonometry, we find that the y component of the dolphin’s velocity is b g v y = v x tan θ = v x tan 35° = 7 .7 m / s tan 35° = 5.4 m / s 16. REASONING a. Here is a summary of the data for the skateboarder, using v0 = 6.6 m/s and θ = 58°: y-Direction Data y ay vy v0y ? −9.80 m/s2 0 m/s +(6.6 m/s)sin 58° = +5.6 m/s t x-Direction Data x ax ? 0 m/s2 vx v0x t +(6.6 m/s)cos 58° = +3.5 m/s 2 We will use the relation v2 = v0 y + 2a y y (Equation 3.6b) to find the skateboarder’s vertical y displacement y above the end of the ramp. When the skateboarder is at the highest point, his vertical displacement y1 above the ground is equal to his initial height y0 plus his vertical displacement y above the end of the ramp: y1 = y0 + y. Next we will use v y = v0 y + a y t 106 KINEMATICS IN TWO DIMENSIONS (Equation 3.3b) to calculate the time t from the launch to the highest point, which, with x = v0 xt + 1 axt 2 (Equation 3.5a), will give us his horizontal displacement x. 2 SOLUTION a. Substituting vy = 0 m/s into Equation 3.6b and solving for y, we find ( 0 m/s ) 2 = 2 v0 y + 2a y y or y= 2 −v0 y 2a y = − ( 5.6 m/s ) ( 2 2 −9.80 m/s2 Therefore, the highest point y1 reached by the y1 = y0 + y = 1.2 m + 1.6 m = +2.8 m above the ground. ) = +1.6 m skateboarder occurs when b. We now turn to the horizontal displacement, which requires that we first find the elapsed time t. Putting vy = 0 m/s into Equation 3.3b and solving for t yields 0 m/s = v0 y + a y t or t= −v0 y ay (1) Given that ax = 0 m/s2, the relation x = v0 xt + 1 axt 2 (Equation 3.5a) reduces to x = v0 x t . 2 Substituting Equation (1) for t then gives the skateboarder’s horizontal displacement: −v0 y − v0 x v0 y − ( +3.5 m/s )( +5.6 m/s ) x = v0 x t = v0 x = = +2.0 m = ay ay −9.80 m/s 17. SSM REASONING The upward direction is chosen as positive. Since the ballast bag is released from rest relative to the balloon, its initial velocity relative to the ground is equal to the velocity of the balloon relative to the ground, so th...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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