Physics Solution Manual for 1100 and 2101

0 ms vx since this table contains three known

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 2 1 = = 12 m 2a y 2a y sin 52° sin 2 52° ___________________________________________________________________________________________ Chapter 3 Problems 115 29. REASONING The vertical displacement y of the ball depends on the time that it is in the air before being caught. These variables depend on the y-direction data, as indicated in the table, where the +y direction is "up." y-Direction Data y ay ? −9.80 m/s2 v0y t 0 m/s ? vy Since only two variables in the y direction are known, we cannot determine y at this point. Therefore, we examine the data in the x direction, where +x is taken to be the direction from the pitcher to the catcher. x-Direction Data x ax +17.0 m 0 m/s v0x 2 t +41.0 m/s vx ? Since this table contains three known variables, the time t can be evaluated by using an equation of kinematics. Once the time is known, it can then be used with the y-direction data, along with the appropriate equation of kinematics, to find the vertical displacement y. SOLUTION Using the x-direction data, Equation 3.5a can be employed to find the time t that the baseball is in the air: x = v0 x t + 1 ax t 2 = v0 x t 2 Solving for t gives t= ( since a x = 0 m/s 2 ) x +17.0 m = = 0.415 s v0 x +41.0 m/s The displacement in the y direction can now be evaluated by using the y-direction data table and the value of t = 0.415 s. Using Equation 3.5b, we have y = v0 y t + 1 a y t 2 = ( 0 m/s ) ( 0.415 s ) + 1 ( −9.80 m/s 2 ) ( 0.415 s ) = −0.844 m 2 2 2 The distance that the ball drops is given by the magnitude of this result, so Distance = 0.844 m . ______________________________________________________________________________ 116 KINEMATICS IN TWO DIMENSIONS 30. REASONING The data for the problem are summarized below. In the tables, we use the symbol v0 to denote the speed with which the ball is thrown and choose upward and to the right as the positive directions. v 30.0° 0 v0y x-Direction Data v0x v0 x t ax vx v0x 183 m v0 cos 30.0° 0 m/s2 30.0° ? 183 m y-Direction Data y ay 0m −9.80 m/s2 v0y t v0 sin 30.0° vy Same as for x direction Note that ax = 0 m/s2, because air resistance is being ignored. In addition, note that y = 0 m, because the football rises and then returns to the same vertical level from which it was launched. Finally, we have used trigonometry to express the components v0x and v0y of the initial velocity in terms of the speed v0 and the 30.0° launch angle. The key here is to remember that the horizontal and vertical parts of the football’s motion can be treated separately, the time for the motion being the same for each. Since air resistance is being ignored, we can apply the equations of kinematics separately to the motions in the x and y directions. SOLUTION Since there is no acceleration in the horizontal direction, motion in that direction is constant-velocity motion, and the horizontal displacement x is simply the initial velocity component v0x times the time t: x = v0 x t = ( v0 cos 30.0° ) t An expression for t can be obtained by considering the motion in the vertical direction. Thus, we use Equation 3.5b from the equations of kinematics and recognize that the displacement y is zero and v0y = v0 sin 30.0°: 1 2 y = v0 y t + a y t 2 or 0 m = ( v0 sin 30.0° ) t + a y t 2 1 2 or Substituting this result for the time into the expression for x gives t= −2v0 sin 30.0° ay Chapter 3 Problems 117 −2v sin 30.0° 2v 2 cos 30.0° sin 30.0° 0 =− 0 x = ( v0 cos 30.0° ) t = ( v0 cos 30.0° ) ay ay − xa y v0 = 2 cos 30.0° sin 30.0° = ( − (183 m ) −9.80 m/s 2 2 cos 30.0° sin 30.0° ) = 45.5 m/s 31. SSM WWW REASONING Once the diver is airborne, he moves in the x direction with constant velocity while his motion in the y direction is accelerated (at the acceleration due to gravity). Therefore, the magnitude of the x component of his velocity remains constant at 1.20 m/s for all times t. The magnitude of the y component of the diver's velocity after he has fallen through a vertical displacement y can be determined from Equation 3.6b: 2 v 2 = v0 y + 2a y y . Since the diver runs off the platform horizontally, v0 y = 0 m/s. Once the x y and y components of the velocity are known for a particular vertical displacement y, the 2 2 speed of the diver can be obtained from v = v x + v y . SOLUTION For convenience, we will take downward as the positive y direction. After the diver has fallen 10.0 m, the y component of his velocity is, from Equation 3.6b, 2 v y = v 0 y + 2 a y y = 0 2 + 2(9.80 m / s 2 )(10.0 m) = 14 .0 m / s Therefore, 2 2 v = v x + v y = (1.20 m / s) 2 + (14.0 m / s) 2 = 14.1 m / s ____________________________________________________________________________________________ 32. REASONING Since we know the launch angle θ = 15.0°, the launch speed v0 can be obtained using trigonometry, which gives the y component of the launch velocity as v0y = v0 sin θ. Solving this equation for v0 requires a value for v0y, which we can obtain from the vertical height of y = 13...
View Full Document

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.

Ask a homework question - tutors are online