Physics Solution Manual for 1100 and 2101

Reasoning a the maximum possible distance that the

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Unformatted text preview: at v0 y = 3.0 m/s . Time required for the ballast to reach the ground can be found by solving Equation 3.5b for t. SOLUTION Using Equation 3.5b, we have 1 2 a y t 2 + v0 y t − y = 0 or 1 2 (–9.80 m / s 2 ) t 2 + ( 3.0 m / s) t − ( −9 .5 m) = 0 This equation is quadratic in t, and t may be found from the quadratic formula. Using the quadratic formula, suppressing the units, and discarding the negative root, we find −3.0 ± (3.0) 2 − 4 ( −4 .90 )( 9 .5) t= = 1.7 s 2 ( −4 .90 ) Chapter 3 Problems 107 ____________________________________________________________________________________________ 18. REASONING a. The maximum possible distance that the ball can travel occurs when it is launched at an angle of 45.0°. When the ball lands on the green, it is at the same elevation as the tee, so the vertical component (or y component) of the ball's displacement is zero. The time of flight is given by the y variables, which are listed in the table below. We designate "up" as the +y direction. y-Direction Data y ay 0m −9.80 m/s v0y t +(30.3 m/s) sin 45.0° = +21.4 m/s ? vy 2 Since three of the five kinematic variables are known, we can employ one of the equations of kinematics to find the time t that the ball is in the air. b. The longest hole in one that the golfer can make is equal to the range R of the ball. This distance is given by the x variables and the time of flight, as determined in part (a). Once again, three variables are known, so an equation of kinematics can be used to find the range of the ball. The +x direction is taken to be from the tee to the green. x-Direction Data x ax R=? 0 m/s vx 2 v0x t +(30.3 m/s) cos 45.0° = +21.4 m/s from part a SOLUTION a. We will use Equation 3.5b to find the time, since this equation involves the three known variables in the y direction: ( ) y = v0 y t + 1 a y t 2 = v0 y + 1 a y t t 2 2 0 m = +21.4 m/s + 1 2 ( −9.80 m/s ) t t 2 Solving this quadratic equation yields two solutions, t = 0 s and t = 4.37 s. The first solution represents the situation when the golf ball just begins its flight, so we discard this one. Therefore, t = 4.37 s . b. With the knowledge that t = 4.37 s and the values for ax and v0x (see the x-direction data table above), we can use Equation 3.5a to obtain the range R of the golf ball. 2 2 1 1 x { = v0 xt + 2 axt = ( +21.4 m/s ) ( 4.37 s ) + 2 ( 0 m/s ) ( 4.37 s ) = 93.5 m 2 =R 108 KINEMATICS IN TWO DIMENSIONS ______________________________________________________________________________ 19. REASONING AND SOLUTION The maximum vertical displacement y attained by a 2 projectile is given by Equation 3.6b ( v 2 = v0 y + 2a y y ) with vy = 0: y y=– 2 v0 y 2a y In order to use Equation 3.6b, we must first estimate his initial speed v0y . When Jordan has reached his maximum vertical displacement, vy = 0, and t = 1.00 s. Therefore, according to Equation 3.3b ( v y = v0 y + a y t ), with upward taken as positive, we find that v0 y = – a y t = – (–9.80 m/s 2 ) (1.00 s) = 9.80 m/s Therefore, Jordan's maximum jump height is y=– (9.80 m/s)2 = 4.90 m 2(–9.80 m/s 2 ) This result far exceeds Jordan’s maximum jump height, so the claim that he can remain in the air for two full seconds is false. ____________________________________________________________________________________________ 20. REASONING The magnitude (or speed) v of the ball’s velocity is related to its x and y velocity components (vx and vy ) by the Pythagorean theorem; 2 v = vx + v 2 y (Equation 1.7). The horizontal component vx of the ball’s velocity never changes during the flight, since, in the absence of air resistance, there is no acceleration in the x direction (ax = 0 m/s2). Thus, vx is equal to the horizontal component v0x of the initial velocity, or v x = v0 x = v0 cos 40.0o . Since v0 is known, vx can be determined. The vertical component vy of the ball’s velocity does change during the flight, since the acceleration in the y direction is that due to gravity (ay = −9.80 m/s2). The relation 2 v 2 = v0 y + 2a y y (Equation 2.6b) may be used to find v 2 , since ay, y, and v0y are known y y (v0y = v0 sin 40.0º). Chapter 3 Problems 109 SOLUTION The speed v of the golf ball just before it lands is 2 v = vx + v2 = y = 2 ( v0 cos 40.0°)2 + v0 y + 2a y y 14 24 3 2 vy ( v0 cos 40.0°) + ( v0 sin 40.0°) 2 2 + 2a y y ( ) = (14.0 m/s ) cos 40.0° + (14.0 m/s ) sin 40.0° + 2 −9.80 m/s2 ( 3.00 m ) = 11.7 m/s 2 2 21. SSM REASONING The time that the ball spends in the air is determined by its vertical motion. The time required for the ball to reach the lake can be found by solving Equation 3.5b for t. The motion of the golf ball is characterized by constant velocity in the x direction and accelerated motion (due to gravity) in the y direction. Thus, the x component of the velocity of the golf ball is constant, and the y component of the velocity at any time t can be found from Equation 3.3b. Once the x and y components of the velocity are known 2 2 for a...
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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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