Questions Chap4-HW1 7. An airplane flying horizontally at a constant speed of 350 km/h over level ground releases a bundle of food supplies. Ignore the effect of the air on the bundle. What are the bundle's initial (a) vertical and (b) horizontal components of velocity? a) voy = 0 km/h, b) vox = 350 km/h, (c) What is its horizontal component of velocity just before hitting the ground? There is no significant horizontal acceleration so vx = 350 km/h, (d) If the airplane's speed were, instead, 450 km/h, would the time of fall be longer, shorter, or the same? Same. Time of fall is determined by vertical acceleration and vertical initial velocity 9. Three paths are shown for a football kicked from ground level. Ignoring the effects of air, rank the paths according to (a) time of flight, Since they all rise to the same height, they all tie; (b) initial vertical velocity component, Since they all rise to the same height all tie (c) initial horizontal velocity component They have different horizontal initial velocities because their ranges are different: 3, 2, 1; (d) initial speed, greatest first: 3, 2, 1 Problems 17. A projectile is fired horizontally from a gun that is 45.0 m above flat ground, emerging from the gun with a speed of 250 m/s. (a) How long does the projectile remain in the air? 1 2 Know that y = y 0 + voy t + at y 2 Choose origin at initial position of ball so yo = 0 and y = -45 m x Since the ball is fired horizontally voy = 0 45 m Therefore 2y 2 (- 45 m ) t= = 1.52 m a (- 9.8 m/s 2 ) = 3.03 s (b) At what horizontal distance from the firing point does it strike the ground? Since there is no horizontal acceleration, x - x0 = vox t = 758 m (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? v y - v0 y = at = - gt Therefore
v y = at = - gt = -29.7 m/s
1
So magnitude is 29.7 m/s.
23. A certain airplane has a speed of 290.0 km/h and is diving at an angle of = 30.0 below the horizontal when the pilot releases a radar decoy. The horizontal distance between the release point and the point where the decoy strikes the ground is d = 700 m. (a) How long is the decoy in the air? Choose axes as shown. Here 0 = 30.0 . The initial speed of the decoy is v0 = 290 km/h = 80.6 m/s. For the horizontal motion: (x - x0 ) 700 x - x 0 = (v cos 0 ) t , t = = = 10.0 s v cos 0 80.6 cos(- 30.0 0 ) (b) How high was the release point? For the vertical motion y0: Therefore y0 = 897 m.
y0
y x
1 1 y - y0 = (v0 sin 0 ) t - gt 2 0 - y0 = (-40.3)(10.0) - (9.80)(10.0) 2 2 2
26. During a tennis match, a player serves the ball at 23.6 m/s, with the center of the ball leaving the racquet horizontally 2.37 m above the court surface. The net is 12 m away and 0.90 m high. When the ball reaches the net, (a) does the ball clear it and (b) what is the distance between the center of the ball and the top of the net? Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00 below the horizontal. When the ball reaches the net, (c) does the ball clear it and (d) what now is the distance between the center of the ball and the top of the net? Hint for 26 a): use the x-component of v to find the time when the ball reaches then net. Then calculate how far the ball drops by the time it has reached the net.
v0
net
y x
a) Choose x and y as shown. The ball will clear the net if its height is greater than the net height when it is at position x = 12 m. To find the height of the ball, first find the time it arrives at the net: x = v0x t = v0t (acceleration along x is zero). t = x/ v0 = (12 m)/(23.6 m/s) = 0.5085 s At this moment, the ball is at a height (above the court) of
y = y 0 + (v sin 0 0 )t - 1 gt 2 = 2.37 m - 1.27 m = 1.10 m is 2
This greater than 0.90 m and so the ball clears the net. .
2
(b) From part a) the height of the ball at the net is 1.10 m, and so the ball is a distance 1.10 0.90 m = 0.20 m above the net. (c) Now the ball is directed downward at 5o to the horizontal so vox = v0cos (-50) = 23.5 m/s and x = v0x t Therefore t = x/ v0x = (12 m)/(23.5 m/s) = 0.510 s At this moment, the ball is at a height (above the court), using voy = vocos , y = y0 + voy t + gt2 = y0 + vocos t + gt2 = 0.05 m This ball cannot clear the net. (d) In part (c), the distance between the top of the net and the center of the ball at t = 0.510 s is 0.90 0.05 = 0.85 m 28. A soccer ball is kicked from the ground with an initial speed of 19.5 m/s at an upward angle of 45. A player 55 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? First find time of flight (since the ball starts and end at the same height, y = 0) 1 (19.5) sin 45.0 y = (v0 sin 0 ) t - gt 2 t = = 2.81 s. 2 (9.80) / 2 Therefore the ball travels a distance x = (v0 cos 0)t = 38.7 m. This means the player must run a distance of 55 m 38.7 m = 16.3 m in 2.81 s Therefore player speed vavg = 16.3/2.81 = 5.8 m/s.
vo
33. In Figure 4-36, a ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.50 s later, at distance d = 25.0 m from the building and at angle = 60.0 with the horizontal. (a) Find h. (Hint: One way is to reverse the motion, as if on videotape.) Hint: 33 a)Know that 2 2 v y - v 0 y = 2a( y - y 0 ) = 2(- g )(- h ) = 2 gh . So need to find the initial and final y velocities to find h. Choose x and y axes are shown. Then xo = 0, x = d, yo = h, y = 0 2 2 Know that v y - v 0 y = 2a( y - y 0 ) = -2 g (0 - h ) = 2 gh
3
0
y
x
Note: we don't know if the ball is thrown above or below horizontal. It will turn out at the end that 0 is negative so ball is thrown below horizontal
So need to find the initial and final y velocities to find h. First vy x - x0 25 x - x 0 = v ox t = v x t = v cos 0 t v= = = 33.3 m/s t cos 0 1.5 cos 60 0
v y = -v sin = -33.3 sin 60 0 = -28.9 m/s
Now find voy v y - v0 y = - gt
v0 y = v y + gt = -28.9 + 9.8 1.5 = -14.2 m/s
2 2 v y - v0 y
28.9 2 - 14.2 2 = = 32.3 m h= Finally 2g 2 9 .8 What are the (b) magnitude (c) angle relative to the horizontal of the velocity at which the ball is thrown? Now need vox : vox = v x = v cos = 16.7 m/s
2 2 v0 = v0 x + v0 y = 16.7 2 + 14.2 2 = 21.9 m/s
tan 0 = v0 y / vox = -14.2 / 16.7 0 = -40.4 0 (d) Is the angle above or below the horizontal? Since 0 is negative, it is below the horizontal
34. A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in Figure 4-37, where t = 0 at the instant the ball is struck. (a) How far does the golf ball travel horizontally before returning to ground level? The horizontal motion takes 5 s. Then x x0 = voxt So need vox Know that vox = vx = constant since there is no acceleration along x. At maximum height (occurs at 2.5 s) know that vy = 0 Therefore st this point v = vx = vox = 19 m/s and x x0 = voxt = 19 m/s 5 s = 95 m. (b) What is the maximum height above ground level attained by the ball? Measure y from ground level so that y0 = 0. Know that y - y 0 = voy t - gt 2 . So need voy. From graph vo = 31 m/s
2 2 Then using vo = vox + voy 2 2 v y = vo - vox = 312 - 19 2 = 24.5 m/s
1 2
Since 192 + v0 2 = 31 m/s (the first point on the graph), then v0 y = 24.5 m/s. y Max height occurs at t = 2.5 s, and so max height is:
y = y 0 + v oy t - 1 2 1 2 gt = 0 + 24.5 m/s 2.5 s - 9.8 m/s 2 (2.5 s ) = 31 m 2 2
4