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4 CHAPTER PROBLEM SOLUTIONS Problem 4. j An airliner is flying at a velocity of 260 m/s , when a wind gust gives it an acceleration of 0.38 + 0.72^ m/s for a period of 24 s. (a) What is its velocity at the end of that time? (b) By what angle has it been deflected from its original course? 2 Solution 2 j j (a) Equation 4-3 gives v = 260 m/s + (0.38 + 0.72^ )(m/s )(24 s) = (269 + 17.3^) m/s. (b) Since v0 is along the x-axis, the angular deflection is just tan (v y= x ) = tan (17.3 269 ) = 3.67 . v = !1 !1 Problem 6. An airplane heads northeastward down a runway, accelerating from rest at the rate of 2.1 m/s . Express the plane's velocity and position at t = 30 s in unit vector notation, using a coordinate system with x-axis eastward and y-axis northward, and with origin at the start of the plane's takeoff roll. 2 Solution Since the acceleration is constant and the airplane starts from rest (v0 = 0) at the origin (r0 = 0), Equations 4-3 and 4-4 2 2 ^ give v(t) = at and r(t) = 1 at . Both vectors are in the NE direction, parallel to a = (2.1 m/s )( cos 45 + j sin 45) = 2 (1.48 m 2)( + ^ ). Thus, v(30 s) = (1.48 m/s2 )(30 s)( + ^ ) = (44.5 m/s)( + ^), and r(30 s) = /s j j j ^ (668 m)( + ^). (Note: ( + j)= 2 is a unit vector in the NE direction.) j 1 2 (1.48 m/s)(30 s) ( + ^ ) = j 2 Problem 13. Figure 4-27 shows a cathode-ray tube, used to display electrical signals in oscilloscopes and other scientific instruments. 9 Electrons are accelerated by the electron gun, then move down the center of the tube at 2.0 ! 10 cm/s . In the 4.2-cmlong deflecting region they undergo an acceleration directed perpendicular to the long axis of the tube. The acceleration "steers" them to a particular spot on the screen, where they produce a visible glow. (a) What acceleration is needed to deflect the electrons through 15, as shown in the figure? (b) What is the shape of an electron's path in the deflecting region? Solution (a) With x-y axes as drawn on Fig. 4-27, the electrons emerge from the deflecting region with velocity v = v 0 + at ^ , j 9 after a time t = x=0 , where x = 4.2 cm and v = 2 ! 10 cm/s. The angle of deflection (direction of v) is 15, so v 0 2 tan 15 = v y= x = at=0 = ax=0 . Thus, a = v0 tan15= = 2.55 ! 101 7 cm/s2 (when values are substituted). (b) Since the v v v2 x acceleration is assumed constant, the electron trajectory is parabolic in the deflecting region. CHAPTER 4 71 Problem 14. You toss an apple horizontally at 8.7 m/s from a height of 2.6 m. Simultaneously, you drop a peach from the same height. How long does each take to reach the ground? Solution The time of flight for either projectile can be determined from the vertical component of the motion, which is the same for both, since v y = 0. Thus, Equation 4-8 gives t = 0 2( y ! y0 )= = g 2( 2.6 m)= (9.8 m/s2 ) = 0.728 s. Problem 18. In a chase scene, a movie stuntman is supposed to run right off the flat roof of one city building and land on another roof 1.9 m lower. If the gap between the buildings is 4.5 m wide, how fast must he run? Solution The horizontal and vertical distances covered by the stuntman are x ! x0 = v0 t and y0 ! y = 2 1 2 gt 2 (since v = v0 , 0x and v y = 0). Eliminating t, one finds v 0 = (x ! x 0 ) g= y 0 ! y) = (4.5 m) (9.8 m/s )= 2( 2(1.9 m) = 7.23 m/s. (Note that 0 Equation 4-9 with !0 = 0 and y0 = 0 provides an equivalent solution.) 72 CHAPTER 4 Problem 22. Derive a general formula for the horizontal distance covered by a projectile launched horizontally at speed v from a 0 height h. Solution This problem is solved by reasoning identical to that in problem 15, where it was shown that x = v0 2 h g. = Problem 15. A carpenter tosses a shingle off a 8.8-m-high roof, giving it an initially horizontal velocity of 11 m/s. (a) How long does it take to reach the ground? (b) How far does it move horizontally in this time? Solution (a) The shingle reaches the ground when y(t) = 0 = y0 ! 1 2 gt2 , or 2(8.8 m) (9.8 m/s2 ) = 1.34 s. t = 2 y0 = g (b) The horizontal displacement is x = v0 t = (11 m /s)(1.34 s) = 14.7 m. CHAPTER 4 73 Problem 28. A rescue airplane is flying horizontally at speed v0 at an altitude h above the ocean, attempting to drop a package of medical supplies to a shipwreck victim in a lifeboat. At what line-of-sight angle ! (Fig. 4-29) should the pilot release the package? Solution The package, when dropped, is a horizontally launched projectile, like that in Problem 15 (the expressions for t and x are the "1 g same). The line-of-sight angle is ! = tan (h x), where x = v 0 2 h= is the horizontal distance at drop time. Thus, = ! = tan "1 gh= 2 . 2v 0 FIGURE 4-29 Problem 28 Solution. Problem 29. At a circus, a human cannonball is shot from a cannon at 35 km/h at an angle of 40. If he leaves the cannon 1.0 m off the ground, and lands in a net 2.0 m off the ground, how long is he in the air? Solution The time of flight can be calculated from Equation 4-8, where the trajectory begins at y0 = 1.0 m and t = 0, ends at y(t) = 2.0 m, and v y = v0 sin !0 = (35 m/3.6 s) sin 40 = 6.25 m/s . The equation is a quadratic, 0 ( y ! y0 ) = 0, with solutions t = [v 0 y 1 2 gt 2 ! v0y t + v 2 ! 2 g( y ! y 0 )]=g = [6.25 m/s 0y (6.25 m/s)2 ! 2(9.8 m/s2 )( 2 m ! 1 m)] (9.8 m/s2 ) = 0.188 s or 1.09 s. The trajectory crosses the height 2.0 m twice, once going up, at the smaller time of flight, and once going down into the net, at the larger time of flight. The latter is the answer to the question asked here. 74 CHAPTER 4 Problem 33. A projectile launched at an angle !0 to the horizontal reaches a maximum height h. Show that its horizontal range is 4h tan !0 . = Solution The intermediate expression for the horizontal range (when the initial and final heights are equal) is x = 2v0 sin! 0 cos ! 0= = 2v0xv0y= (see the equation before Equation 4-10). The components of the initial velocity are related by g g 2 v y=0x = tan !0 . The maximum height, h = y max ! y0 , can be found from Equation 2-11 (when vy = 0) or v2 y = 2 gh. 0 v 0 Then x = 2v0 yv0x= = 2v0 y (v0 y= !0 )= = 2(2gh)= tan !0 = 4h= !0 . (This result reflects a classical geometrical property g tan g g tan of the parabola, namely, that the latus rectum is four times the distance from vertex to focus.) Problem 34. You're 5.0 m from the left-hand wall of the house shown in Fig. 4-30, and you want to throw a ball to a friend 5.0 m from the right-hand wall. (a) What is the minimum speed that will allow the ball to clear the roof? (b) At what angle should it be thrown? Assume the throw and catch both occur 1.0 m above the ground. FIGURE 4-30 Problem 34. Solution Since the trajectory is symmetrical (begins and ends at the same height), one can use the result of the previous problem with "1 x) h = y max ! y0 = 6 m ! 1 m = 5 m and horizontal range x = 5 m+ 6 m + 5 m = 16 m. Then (b) !0 = tan (4h= = 51.3 , and (a) v 0 = v 0y = ! 0 = sin 2 gh= ! 0 = 12.7 m/s. sin Problem 42. How fast would a car have to round a turn 75 m in radius in order for its acceleration be to numerically equal to that of gravity? Solution For circular motion with constant speed, ar = v = or r 2 v = ar r = (9.8 m/s 2 )( 75 m) = 27.1 m/s = 97.6 km/h = 60.7 mi h. / CHAPTER 4 75 Problem 46. A 12-in-diameter circular saw blade rotates at 3500 revolutions per minute. What is the acceleration of one of the saw teeth? Compare with the acceleration of gravity. Solution If the saw blade rotates at 3500 rpm, a point on its circumference has a linear speed of v = (3500)! (12 in)= s = 183 ft/s. 60 The (radial) acceleration is ar = v = = (183 ft/s) = ft= = 6.72 ! 10 ft/s ' 2 ! 10 g, where g = 32.2 ft/s . r (6 12) Alternatively, one could use Equation 4-12 with period equal to (3500) !1 2 2 4 2 3 2 min . Problem 54. A car moving at 65 km/h enters a curve that describes a quarter turn of radius 120 m. The driver gently applies the 2 brakes, giving a constant tangential deceleration of magnitude 0.65 m/s . Just before emerging from the turn, what are (a) the magnitude of the car's acceleration and (b) the angle between the acceleration vector and the direction of motion? Solution The car's tangential acceleration is constant, a = !0.65 m/ s along the direction of motion, so Equation 2-11 can be used to determine the speed at the end of the turn, v = v0 + 2at s. Here, s = 2! r= is the linear distance around the quarter turn. 4 Then the centripetal acceleration is ac = v = = (v 0 + ! rat )= = (65 m= s) = r r 3.6 120 m + ! ("0.65 m/s ) = 0.675 m/ s . (a) The magnitude of the total acceleration is "1 2 2 2 2 2 2 2 2 a2 + at2 = 0.937 m/ s2. (b) The angle of the total acceleration with respect to c the tangent to the curve is ! = tan (ac= t ) = 134. (This angle is in the second quadrant because the tangential acceleration a is opposite to the direction of motion.) Problem 57. If you can throw a stone straight up to a height of 16 m, how far could you throw it horizontally over level ground? Assume the same throwing speed and optimum launch angle. Solution To throw an object vertically to a maximum height of h = 16 m = y max ! y0 requires an initial speed of v = 0 2 g( ymax ! y0 ) = 2 gh . With this value of v and the optimum launch angle !0 = 45, Equation 4-10 gives a maximum 0 horizontal range on level ground of x = v0 = = 2h = 32 m. (The maximum horizontal range on level ground is twice the g maximum height for vertical motion with the same initial speed. This result holds in the approximation of constant g and no air resistance.) 2 76 CHAPTER 4 Problem 63. You toss a chocolate bar to your hiking companion located 8.6 m up a 39 slope, as shown in Fig. 4-36. Determine the initial velocity vector so that the chocolate bar will reach your friend moving horizontally. FIGURE 4-36 Problem 63. Solution The candy bar moves horizontally only at the apex of its trajectory (where vx = v0x and vy = 0). Thus, ymax ! y0 = 2 (8.6 m) sin 39 = 5.41 m, and v y = 2 g( ymax ! y0 ) = 2(9.8 m/s )( 5.41 m) = 10.3 m/s (see Equation 2-11). The time to 0 t reach the apex is t = v0y= so v 0x = (x ! x0 )= = (x ! x0)g=v 0y (see Equations 4-6 and 4-7). The horizontal distance from g, apex to origin is x ! x0 = (8.6 m) cos 39 = 6.68 m, so v 0x = (6.68 m)(9.8 m/s )= (10.3 m/s) = 6.36 m/s. v0 can be 2 ^ ) m/s, or by its magnitude v 2 + v 0 y = 12.1 m/s and direction ! = expressed in unit vector notation as (6.36 + 10.3j 0x 2 tan!1 (v y=0x ) = 58.3 (CCW from the x-axis). 0 v CHAPTER 4 77 Problem 69. A monkey is hanging from a branch a height h above the ground. A naturalist stands a horizontal distance d from a point directly below the monkey. The naturalist aims a tranquilizer dart directly at the monkey, but just as she fires the monkey 2 2 2h lets go. Show that the dart will nevertheless hit the monkey, provided its initial speed exceeds (d + h )g= . Solution Gravity accelerates the dart and the monkey equally, so both fall the same vertical distance from the point of aim (the monkey's original position) resulting in a hit, provided the initial speed of the dart is sufficient to reach the monkey before the monkey reaches the ground. To prove this assertion, let the dart be fired from ground level ( y = 0) with speed v0 and "1 direction !0 = tan (h d) (line of sight from naturalist N to monkey M), while the monkey drops from height h at t = 0. The = vertical height of each is ymonkey = h ! 1 2 gt2 and ydart = v 0 yt ! 1 2 2 2 gt 2 , where v 0 y = v 0 sin ! 0 = v 0 h= d + h . (The term ! 1 2 gt2 represents the effect of gravity, which appears the same way in both y-coordinate equations.) The dart strikes the monkey when ymonkey = ydart , which implies h = v0 yt , or t = required for the monkey to fall to the ground, which is d 2 + h2 =v 0 . This must be less than the time d 2 + h2 = 0 < v 2h=g or 2h= (from ymonkey = 0). Thus g g(d + h )= (This condition can also be understood from the horizontal range formula, Equation 4-10. The 2h. 2 2 2 range of the dart has to be greater than the horizontal distance to the monkey, d < v 0 sin 2! 0=g = v 0 2hd= + h ) g. (d2 ) v0 > 2 2 Problem 69 Solution. 78 CHAPTER 4 Problem 70. A child tosses a ball over a flat-roofed house 3.2 m high and 7.4 m wide, so it just clears the corners on both sides, as shown in Fig. 4-37. If the child stands 2.1 m from the wall, what are the ball's initial speed and launch angle? Assume the ball is launched essentially from ground level. Solution Choose the coordinate system shown in Fig. 4-37, such that the four given points on the trajectory are the origin (x0 , y0 ) = (0, 0), the first corner (x1, y1 ) = (2.1 m, 3.2 m), the second corner (x2 , y2 ) = (9.5 m, 3.2 m), and ground level at the right (x3 , y3 ) = (11.6 m, 0). Equation 4-9 for the trajectory, evaluated at any two points other than the origin, provides two equations which can be solved for v and !0 . Before substituting values, we may divide Equation 4-9 by x, and let 0 2v2 2 a = tan ! 0 and b = g= 0 cos !0 . For example, selecting the two corner points, we obtain y1=x1 = a ! x1 b, and y2= 2 = a ! x2 b, with solutions (x2 ! x1)a = ( x2 y1 =x1) ! (x1 y2 =x2 ), and (x2 ! x1)b = ( y1 =x1) ! (y2 =x2). In terms of a x and b, !0 = tan "1 a , and v 0 = g(1 + a2 )= b, so with the given numerical values, one finds 2 1 % " 9.5 2.1% " 1 7.4a = $ ! ' 3.2, and 7.4b = $ ! ' 3.2, # 2.1 9.5& # 2.1 m 9.5 m& or !0 = tan "1 1.86 = 61.7, and v 0 = (9.8 m/s2)( 1 + 1.862 )= 2(0.160 m!1) = 11.7 m/s. FIGURE 4-37 Problem 70 Solution. Problem 72. In your calculus class, you may have learned that you can find the maximum or minimum of a function by differentiating and setting the result to zero. Do this for Equation 4-10, differentiating with respect to !0 , and thus verify that the maximum range occurs for !0 = 45. Solution The derivative of Equation 4-10 with respect to !0 is dx=d!0 = 2(v 0 =g) cos 2! 0 . This is zero when 2! 0 = 90, or !0 = 45 as stated. (This is the only maximum, since launch angles are restricted to the range 0 < !0 < 90, and d x=d!0 < 0.) 2 2 2 ... View Full Document