Physics Solution Manual for 1100 and 2101

5 m sg b 5 m sg v f i tan f 5 m s i 377 691 g

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Unformatted text preview: tion gives 2 ( −2.0 m ) 2y = ( 5.3 m/s ) = 3.4 m ay −9.8 m/s2 D = v0 x t = v0 x 25. REASONING The magnitude and direction of the initial velocity v0 can be obtained using the Pythagorean theorem and trigonometry, once the x and y components of the initial velocity v0x and v0y are known. These components can be calculated using Equations 3.3a and 3.3b. SOLUTION Using Equations 3.3a and 3.3b, we obtain the following results for the velocity components: b= c h565 sg 893.5 m / s − a t = 4816 m / s − c.30 m / s h s g 691.5 m / s 7 565 b= v 0 x = v x − a x t = 3775 m / s − 5.10 m / s 2 v0 y = v y 2 y Using the Pythagorean theorem and trigonometry, we find 893 691 b .5 m / sg+ b .5 m / sg = v F I = tan F .5 m / s I = 37.7 ° 691 G J G .5 m / s J K v 893 HK H 2 2 v0 = v0 x + v0 y = θ = tan −1 2 2 1130 m / s −1 0y 0x ____________________________________________________________________________________________ 26. REASONING We begin by considering the flight time of the ball on the distant planet. Once the flight time is known, we can determine the maximum height and the range of the ball. The range of a projectile is proportional to the time that the projectile is in the air. Therefore, the flight time on the distant plant 3.5 times larger than on earth. The flight time can be found from Equation 3.3b ( v y = v0 y + a y t ). When the ball lands, it is at the same level as the tee; therefore, from the symmetry of the motion v y = – v0 y . Taking upward and to the right as the positive directions, we find that the flight time on earth would be t= v y − v0 y ay = −2v0 y ay = −2v0 sin θ ay = −2(45 m/s) sin 29° –9.80 m/s 2 = 4.45 s Chapter 3 Problems 113 Therefore, the flight time on the distant planet is 3.5 × (4.45 s) = 15.6 s. From the symmetry of the problem, we know that this is twice the amount of time required for the ball to reach its maximum height, which, consequently, is 7.80 s. SOLUTION a. The height y of the ball at any instant is given by Equation 3.4b as the product of the average velocity component in the y direction 1 2 v e 0y j + v y and the time t: y = 1 2 v e 0y j + vy t . Since the maximum height H is reached when the final velocity component in the y direction is zero (vy = 0 m/s), we find that 1 1 1 H = 2 v 0 y t = 2 v 0 (sin 29 ° ) t = 2 (45 m / s) (sin 29 ° ) (7.80 s) = 85 m b. The range of the ball on the distant planet is x = v 0 x t = v 0 (cos 29 ° ) t = ( 45 m / s) (cos 29 ° ) (15.6 s) = 610 m 27. SSM REASONING AND SOLUTION The water exhibits projectile motion. The x component of the motion has zero acceleration while the y component is subject to the acceleration due to gravity. In order to reach the highest possible fire, the displacement of the hose from the building is x, where, according to Equation 3.5a (with ax = 0 m/s2), x = v0 xt = (v0 cos θ )t with t equal to the time required for the water the reach its maximum vertical displacement. The time t can be found by considering the vertical motion. From Equation 3.3b, v y = v0 y + a y t When the water has reached its maximum vertical displacement, vy = 0 m/s. Taking up and to the right as the positive directions, we find that t= −v0 y ay = −v0 sin θ ay and −v sin θ x = (v0 cos θ ) 0 ay Therefore, we have x=− 2 v0 cos θ sin θ ay =− (25.0 m/s) 2cos 35.0° sin 35.0° = 30.0 m –9.80 m/s 2 114 KINEMATICS IN TWO DIMENSIONS 28. REASONING When the ball is thrown straight up with an initial speed v0, the maximum 2 height y that it reaches can be found by using with the relation v 2 = v0 y + 2 a y y (Equation y 3.6b). Since the ball is thrown straight up, v0y = v0, where v0 is the initial speed of the ball. Also, the speed of the ball is momentarily zero at its maximum height, so vy = 0 m/s at that point. The acceleration ay is that due to gravity, so the only unknown besides y is the initial speed v0 of the ball. To determine v0 we will employ Equation 3.6b a second time, but now it will be applied to the case where the ball is thrown upward at an angle of 52º above the horizontal. In this case the maximum height reached by the ball is y1 = 7.5 m, the initial speed in the y direction is v0y = v0 sin 52º, and the y-component of the speed at the maximum height is vy = 0 m/s. 2 SOLUTION We will start with the relation v 2 = v0 y + 2 a y y (Equation 3.6b) to find the y maximum height y that the ball attains when it is thrown straight up. Solving this equation for y, and substituting in v0y = v0 and vy = 0 m/s gives y=− 2 v0 2a y (1) 2 To determine v0, we now apply the equation v 2 = v0 y + 2 a y y to the situation where the ball y is thrown upward at an angle of 52º relative to the horizontal. In this case we note that vy = v0 sin 52º, vy = 0 m/s, and y = y1 (the maximum height of 7.5 m reached by the ball). 2 Solving for v0 , we find v 2 = ( v0 sin 52° ) + 2a y y1 y 2 or 2 v0 = −2a y y1 sin 2 52° 2 Substituting this expression for v0 into Equation (1) gives y=− 2 v0 −2a y y1 2 y 7.5 m = − sin 52°...
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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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