Physics Solution Manual for 1100 and 2101

7 ms tan 75 vx here we have used the fact that the

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Unformatted text preview: .5 m by using Equation 3.6b from the equations of kinematics. SOLUTION From Equation 3.6b we have 2 2 v y = v0 y + 2a y y 0 b m / s g= v 2 2 0y b h c + 2 −9 .80 m / s 2 13.5 m g or c Using trigonometry, we find v0 = v0 y sin 15.0 ° = c b h 2 9 .80 m / s 2 13.5 m sin 15.0 ° b h v 0 y = 2 9 .80 m / s 2 13.5 m g= 62.8 m / s g 118 KINEMATICS IN TWO DIMENSIONS ____________________________________________________________________________________________ y 33. REASONING The drawing at the right shows the velocity vector v of the water at a point below the top of the falls. The components of the velocity are also shown. The angle θ is given by tan θ = v y / v x , so that x θ vy v θ v y = – vx tan θ = –(2.7 m/s) tan 75° vx Here we have used the fact that the horizontal velocity component vx remains unchanged at its initial value of 2.7 m/s as the water falls. Knowing the y component of the velocity, we 2 can use Equation 3.6b, ( v 2 = v0 y + 2a y y ) to find the vertical distance y. y SOLUTION Taking v0 y = 0 m/s, and taking upward as the positive direction, we have from Equation 3.5b that y= v2 y 2a y = –( 2 .7 m / s) tan 75° 2 2(–9.80 m / s 2 ) = –5.2 m Therefore, the velocity vector of the water points downward at a 75° angle below the horizontal at a vertical distance of 5.2 m below the edge. ____________________________________________________________________________________________ 34. REASONING In the absence of air resistance, there is no acceleration in the x direction (ax = 0 m/s2), so the range R of a projectile is given by R = v0x t, where v0x is the horizontal component of the launch velocity and t is the time of flight. We will show that v0x and t are 2 each proportional to the initial speed v0 of the projectile, so the range is proportional to v0 . Since v0x = v0 cos θ, where θ is the launch angle, we see that v0x is proportional to v0. To show that the time of flight t is also proportional to the launch speed v0, we use the fact that, for a projectile that is launched from and returns to ground level, the vertical displacement is y = 0 m. Using the relation y = v0 y t + 1 a y t 2 (Equation 3.5b), we have 2 0 m = v0 y t + 1 a y t 2 2 or 0 m = v0 y + 1 a y t 2 or t= −2v0 y ay Thus, the flight time is proportional to the vertical component of the launch velocity v0y, which, in turn, is proportional to the launch speed v0 since v0y = v0 sin θ. Chapter 3 Problems 119 SOLUTION The range of a projectile is given by R = v0x t and, since both v0x and t are 2 proportional to v0, the range is proportional to v0 . The given range is 23 m. When the launch speed doubles, the range increases by a factor of 22 = 4, since the range is proportional to the square of the speed. Thus, the new range is bg R = 4 23 m = 92 m 35. REASONING The rocket will clear the top of the wall by an amount that is the height of the rocket as it passes over the wall minus the height of the wall. To find the height of the rocket as it passes over the wall, we separate the rocket’s projectile motion into its horizontal and vertical parts and treat each one separately. From the horizontal part we will obtain the time of flight until the rocket reaches the location of the wall. Then, we will use this time along with the acceleration due to gravity in the equations of kinematics to determine the height of the rocket as it passes over the wall. SOLUTION velocity We begin by finding the horizontal and vertical components of the launch b g sin 60.0 ° = b .0 m / s g 60.0 ° 75 sin v 0 x = v 0 cos 60.0 ° = 75.0 m / s cos 60.0 ° v0 y = v0 Using v0x, we can obtain the time of flight, since the distance to the wall is known to be 27.0 m: 27 .0 m 27 .0 m t= = = 0.720 s v0 x 75.0 m / s cos 60.0 ° b g The height of the rocket as it clears the wall can be obtained from Equation 3.5b, in which we take upward to be the positive direction. The amount by which the rocket clears the wall can then be obtained: 1 y = v0 y t + 2 a y t 2 b g b g b gc 1 y = 75.0 m / s sin 60.0 ° 0.720 s + 2 −9 .80 m / s 2 b h0.720 sg= 44.2 m 2 clearance = 44 .2 m − 11.0 m = 33.2 m _____________________________________________________________________________________________ 120 KINEMATICS IN TWO DIMENSIONS 36. REASONING We will treat the horizontal and vertical y parts of the motion separately. The directions upward v0x and to the right are chosen as the positive directions in the drawing. Ignoring air resistance, we can apply the equations of kinematics to the vertical part of the motion. The data for the vertical motion are x summarized in the following tables. Note that the initial velocity component v0y is zero, because the bullets are fired horizontally. The vertical component y of the bullets’ displacements are entered in the tables with minus signs, because the bullets move downward in the negative y direction. The times of flight tA and tB have been identified in the tables as a matter of convenience. y-Direction Data, First Shot y ay −HA −9.80 m/s2 y-Direction Data, Second Shot v0y t y ay 0 m/s tA −HB −9.80 m/s2 vy vy...
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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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