Physics Solution Manual for 1100 and 2101

We are given that the final velocity makes an angle

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Unformatted text preview: omponent y of the marble’s displacement is entered in the table as −H, where H is the height we seek. The minus sign is included, because y-Direction Data the marble moves downward in the negative y direction. The vertical y t ay vy v0y component vy of the final velocity is checked as an important variable in 0 m/s −H √ −9.80 m/s2 the table, because we are given the angle that the final velocity makes with respect to the horizontal. Ignoring air resistance, we apply the equations of kinematics. With the data indicated in the table, Equation 3.6b becomes v2 2 y 2 v 2 = v0 y + 2a y y = ( 0 m/s ) + 2a y ( − H ) or H = − (1) y 2a y 128 KINEMATICS IN TWO DIMENSIONS SOLUTION To use Equation (1), we need to determine the vertical component vy of the final velocity. We are given that the final velocity makes an angle of 65° with respect to the horizontal, as the inset in the drawing shows. Thus, from trigonometry, it follows that −v y tan 65° = v y = −vx tan 65° or vx or v y = −v0 x tan 65° where the minus sign is included, because vy points downward in the negative y direction. In the absence of air resistance, there is no acceleration in the x direction, and the horizontal component vx of the final velocity is equal to the initial value v0x. Substituting this result into Equation (1) gives H =− v2 y 2a y ( −v0 x tan 65°)2 = − − (15 m/s ) tan 65° =− ( 2 −9.80 m/s 2 2a y ) 2 = 53 m 45. REASONING Using the data given in the problem, we can find the maximum flight time t 1 2 of the ball using Equation 3.5b ( y = v0 y t + a y t 2 ). Once the flight time is known, we can use the definition of average velocity to find the minimum speed required to cover the distance x in that time. SOLUTION Equation 3.5b is quadratic in t and can be solved for t using the quadratic formula. According to Equation 3.5b, the maximum flight time is (with upward taken as the positive direction) t= 2 – v0 y ± v0 y – 4 2 = ( )a ( )a 1 2 1 2 y (– y) y – (15.0 m/s ) sin 50.0° ± = 0.200 s and = 2 – v0 y ± v0 y + 2a y y ay (15.0 m/s ) sin 50.0° +2(–9.80 m/s 2 ) (2.10 m) 2 –9.80 m/s 2 2.145 s where the first root corresponds to the time required for the ball to reach a vertical displacement of y = +2.10 m as it travels upward, and the second root corresponds to the time required for the ball to have a vertical displacement of y = +2.10 m as the ball travels upward and then downward. The desired flight time t is 2.145 s. During the 2.145 s, the horizontal distance traveled by the ball is x = vx t = (v0 cos θ ) t = [ (15.0 m/s) cos 50.0°] (2.145 s) = 20.68 m Chapter 3 Problems 129 Thus, the opponent must move 20.68 m –10.0 m = 10.68 m in 2.145 s – 0.30 s = 1.845 s . The opponent must, therefore, move with a minimum average speed of v min = 10.68 m = 5.79 m / s 1.845 s 46. REASONING We will treat the horizontal and vertical parts of the motion separately. The range R is the product of the horizontal component of the initial velocity v0x and the time of flight t. The time of flight can be obtained from the vertical part of the motion by using Equation 3.5b y e= v 0y 1 t + 2 ayt 2 j and the fact that the displacement y in the vertical direction is zero, since the projectile is launched from and returns to ground level. The expression for the range obtained in this way can then be applied to obtain the desired launch angle for doubling the range. SOLUTION The range of the projectile is c h R = v 0 x t = v 0 cos θ t Using Equation 3.5b, we obtain the time of flight as 1 y = 0 = v0 y t + 2 a y t 2 or t=− 2v0 y ay =− 2 v 0 sin θ ay Substituting this expression for t into the range expression gives F2 v sin θ I = − 2 v − R = c cos θ h v Ga J H K 0 2 0 cos θ sin θ 0 ay y =− 2 v 0 sin 2θ ay where we have used the fact that 2 cos θ sin θ = sin 2θ. We can now apply this expression for the range to the initial range R1 for θ1 and the range R2 for θ2: R1 = − 2 v 0 sin 2θ 1 ay or R2 = − 2 v 0 sin 2θ 2 ay Dividing these two expressions gives R2 R1 = c h − c sin 2θ ha v / 2 − v 0 sin 2θ 2 / a y 2 0 1 y = sin 2θ 2 sin 2θ 1 =2 where we have used the fact that R2/R1 = 2. Since θ1 = 12.0°, we find that 130 KINEMATICS IN TWO DIMENSIONS bg sin 2θ 2 = 2 .00 sin 2θ 1 = 2 .00 sin 2 12 .0 ° = 0.813 θ2 = sin −1 0.813 = 27 .2 ° 2 .00 47. SSM REASONING AND SOLUTION In the absence of air resistance, the bullet exhibits projectile motion. The x component of the motion has zero acceleration while the y component of the motion is subject to the acceleration due to gravity. The horizontal distance traveled by the bullet is given by Equation 3.5a (with ax = 0 m/s2): x = v0 xt = (v0 cos θ )t with t equal to the time required for the bullet to reach the target. The time t can be found by considering the vertical motion. From Equation 3.3b, v y = v0 y + a y t When the bullet reaches the target, v y = −v0 y . Assuming that up and to the right are the positive directions, we have t= −2v0 y ay =...
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