(a) Taking the derivative of the position vector with respect to time, we have, in SI units (m/s),

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1 Chapter 4 Student Solutions Manual. We apply Eq. 4- and Eq (a) Taking the deriatie of the position ector with respect to time, we hae, in SI units (m/s), d ˆ = (i + 4t ˆj + tk) ˆ = 8tˆj + k ˆ. dt (b) Taking another deriatie with respect to time leads to, in SI units (m/s ), d a = (8tˆj + k) ˆ = 8ˆj. dt 7. Constant acceleration in both directions (x and y) allows us to use Table - for the motion along each direction. This can be handled indiidually (for Δx and Δy) or together with the unit-ector notation (for Δr). Where units are not shown, SI units are to be understood. (a) The elocity of the particle at any time t is gien by = + at, where is the initial elocity and a is the (constant) acceleration. The x component is x = x + a x t = 3..t, and the y component is y = y + a y t =.5t since y =. When the particle reaches its maximum x coordinate at t = t m, we must hae x =. Therefore, 3..t m = or t m = 3. s. The y component of the elocity at this time is y =.5(3.) =.5 m/s; this is the only nonzero component of at t m. (b) Since it started at the origin, the coordinates of the particle at any time t are gien by r = t + at. At t = t m this becomes ( )( ) ( )( ) r = 3.i ˆ 3. +.ˆi.5ˆj 3. = (4.5ˆi.5ˆ j) m. 9. The initial elocity has no ertical component only an x component equal to +. m/s. Also, y = +. m if the water surface is established as y =. (a) x x = x t readily yields x x =.6 m. (b) Using y y = t gt, we obtain y = 6.86 m when t =.8 s and y =. y

2 (c) Using the fact that y = and y =., the equation y y = t gt leads to y t = (. m)/9.8 m/s =.43 s. During this time, the x-displacement of the dier is x x = (. m/s)(.43 s) =.86 m. 3. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at ground leel directly below the release point. We write θ = 37. for the angle measured from +x, since the angle gien in the problem is measured from the y direction. We note that the initial speed of the projectile is the plane s speed at the moment of release. (a) We use Eq. 4- to find : y y = ( sin θ ) t gt 73 m = sin( 37. )(5. s) (9.8 m/s )(5. s) which yields = m/s. (b) The horizontal distance traeled is x = tcos θ = ( m/s)(5. s)cos( 37. ) = 86 m. (c) The x component of the elocity (just before impact) is x = cosθ = ( m/s)cos( 37. ) = 6 m/s. (d) The y component of the elocity (just before impact) is y = sin θ gt = ( m/s) sin ( 37. ) (9.8 m/s )(5. s) = 7 m/s. 39. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at the end of the rifle (the initial point for the bullet as it begins projectile motion in the sense of 4-5), and we let θ be the firing angle. If the target is a distance d away, then its coordinates are x = d, y =. The projectile motion equations lead to d = t cos θ and = t sinθ gt. Eliminating t leads to sinθ cosθ gd =. Using sinθ cosθ = sinbθ g, we obtain gd (9.8 m/s )(45.7 m) sin ( θ) = gd sin( θ) = = (46 m/s) 3 which yields sin( θ) =. and consequently θ =.66. If the gun is aimed at a point a distance aboe the target, then tan θ = d so that = d tan θ = (45.7 m) tan(.66 ) =.484 m = 4.84 cm.

3 47. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at ground leel directly below impact point between bat and ball. The Hint gien in the problem is important, since it proides us with enough information to find directly from Eq (a) We want to know how high the ball is from the ground when it is at x = 97.5 m, which requires knowing the initial elocity. Using the range information and θ = 45, we use Eq. 4-6 to sole for : ( )( ) gr 9.8 m/s 7 m = = = 3.4 m/s. sin θ Thus, Eq. 4- tells us the time it is oer the fence: x 97.5 m t = = = 4.6 s. cos θ 3.4 m/s cos 45 ( ) At this moment, the ball is at a height (aboe the ground) of b g m y = y + sin θ t gt = 988. which implies it does indeed clear the 7.3 m high fence. (b) At t = 4.6 s, the center of the ball is 9.88 m 7.3 m =.56 m aboe the fence. 5. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. The coordinate origin is at the point where the ball is kicked. We use x and y to denote the coordinates of ball at the goalpost, and try to find the kicking angle(s) θ so that y = 3.44 m when x = 5 m. Writing the kinematic equations for projectile motion: x = y = t gt cos θ, sin θ, we see the first equation gies t = x/ cos θ, and when this is substituted into the second the result is y = x tan θ gx cos. θ One may sole this by trial and error: systematically trying alues of θ until you find the two that satisfy the equation. A little manipulation, howeer, will gie an algebraic solution: Using the trigonometric identity / cos θ = + tan θ, we obtain

4 gx tan gx θ x tan θ + y + = which is a second-order equation for tan θ. To simplify writing the solution, we denote ( )( ) ( ) c= gx / = 9.8 m/s 5 m / 5 m/s = 9.6m. Then the second-order equation becomes c tan θ x tan θ + y + c =. Using the quadratic formula, we obtain its solution(s). ( ) ( )( ) ( ) x± x 4 y+ c c 5 m ± (5 m) m m 9.6 m tan θ = =. c 9.6 m The two solutions are gien by tan θ =.95 and tan θ =.65. The corresponding (firstquadrant) angles are θ = 63 and θ = 3. Thus, (a) The smallest eleation angle is θ = 3, and (b) The greatest eleation angle is θ = 63. If kicked at any angle between these two, the ball will trael aboe the cross bar on the goalposts. 53. We denote h as the height of a step and w as the width. To hit step n, the ball must fall a distance nh and trael horizontally a distance between (n )w and nw. We take the origin of a coordinate system to be at the point where the ball leaes the top of the stairway, and we choose the y axis to be positie in the upward direction. The coordinates of the ball at time t are gien by x = x t and y = gt (since y = ). We equate y to nh and sole for the time to reach the leel of step n: The x coordinate then is t nh =. g nh n(.3 m) x = x = (.5 m/s) = (.39 m) n. g 9.8 m/s The method is to try alues of n until we find one for which x/w is less than n but greater than n. For n =, x =.39 m and x/w =.5, which is greater than n. For n =, x =.437 m and x/w =.5, which is also greater than n. For n = 3, x =.535 m and x/w =.64. Now, this is less than n and greater than n, so the ball hits the third step.

5 67. To calculate the centripetal acceleration of the stone, we need to know its speed during its circular motion (this is also its initial speed when it flies off). We use the kinematic equations of projectile motion (discussed in 4-6) to find that speed. Taking the +y direction to be upward and placing the origin at the point where the stone leaes its circular orbit, then the coordinates of the stone during its motion as a projectile are gien by x = t and y gt (since y = ). It hits the ground at x = m and y =. m. = Formally soling the second equation for the time, we obtain t = y/g, which we substitute into the first equation: b g b g g 98. m/s = x = m = 5. 7 m/s. y. m Therefore, the magnitude of the centripetal acceleration is b g 57. m/s a = = = 6 m/s. r 5. m 75. Relatie to the car the elocity of the snowflakes has a ertical component of 8. m/s and a horizontal component of 5 km/h = 3.9 m/s. The angle θ from the ertical is found from which yields θ = 6. h 3.9 m/s tanθ = = = m/s 77. Since the raindrops fall ertically relatie to the train, the horizontal component of the elocity of a raindrop is h = 3 m/s, the same as the speed of the train. If is the ertical component of the elocity and θ is the angle between the direction of motion and the ertical, then tan θ = h /. Thus = h /tan θ = (3 m/s)/tan 7 =.9 m/s. The speed of a raindrop is = + = ( 3 m/s) + (. 9 m/s) = 3 m/s. h 9. We adopt the positie direction choices used in the textbook so that equations such as Eq. 4- are directly applicable. (a) With the origin at the firing point, the y coordinate of the bullet is gien by y = gt. If t is the time of flight and y =.9 m indicates where the bullet hits the target, then

6 ( ).9 m t = = 6. s. 9.8 m/s (b) The muzzle elocity is the initial (horizontal) elocity of the bullet. Since x = 3 m is the horizontal position of the target, we hae x = t. Thus, x 3 m = = = t 6.3 s 4.8 m/s. 7. (a) Eq. -5 can be applied to the ertical (y axis) motion related to reaching the maximum height (when t = 3. s and y = ): y max y = y t gt. With ground leel chosen so y =, this equation gies the result y max = g(3. s) = 44 m. (b) After the moment it reached maximum height, it is falling; at t =.5 s, it will hae fallen an amount gien by Eq. -8 y fence y max = ()(.5 s) g(.5 s) which leads to y fence = 3 m. (c) Either the range formula, Eq. 4-6, can be used or one can note that after passing the fence, it will strike the ground in.5 s (so that the total "fall-time" equals the "rise-time"). Since the horizontal component of elocity in a projectile-motion problem is constant (neglecting air friction), we find the original x-component from 97.5 m = x (5.5 s) and then apply it to that final.5 s. Thus, we find x = 7.7 m/s and that after the fence Δx = (7.7 m/s)(.5 s) = 8.9 m.. Since the x and y components of the acceleration are constants, we can use Table - for the motion along both axes. This can be handled indiidually (for Δx and Δy) or together with the unit-ector notation (for Δr). Where units are not shown, SI units are to be understood. (a) Since r =, the position ector of the particle is (adapting Eq. -5) ( ˆ) ( ˆ ˆ) ( ) ˆ ( ) r = t + at = 8. j t + 4. i +. j t =.t i + 8. t +.t ˆ j.

7 Therefore, we find when x = 9 m, by soling.t = 9, which leads to t = 3.8 s. The y coordinate at that time is y = (8. m/s)(3.8 s) + (. m/s )(3.8 s) = 45 m. (b) Adapting Eq. -, the elocity of the particle is gien by Thus, at t = 3.8 s, the elocity is = + at. ( )( ) = (8. m/s) ˆj + (4. m/s ) ˆi + (. m/s ) ˆj 3.8 s = (5. m/s) ˆi + (5.6 m/s) ˆ j which has a magnitude of = + = (5. m/s) + (5.6 m/s) = m/s. x y. On the one hand, we could perform the ector addition of the displacements with a ector-capable calculator in polar mode ((75 37 ) + (65 9 ) = (63 8 )), but in keeping with Eq. 3-5 and Eq. 3-6 we will show the details in unit-ector notation. We use a standard coordinate system with +x East and +y North. Lengths are in kilometers and times are in hours. (a) We perform the ector addition of indiidual displacements to find the net displacement of the camel. Δ r ˆ ˆ = (75 km)cos(37 ) i + (75 km) sin(37 ) j Δr ˆ =( 65 km) j Δ r =Δr + Δr = (6 km) ˆi ( km )j ˆ. If it is desired to express this in magnitude-angle notation, then this is equialent to a ector of length Δ r = (6 km) +( km) = 63 km. (b) The direction of Δr is θ = tan [( km)/(6 km)] = 8, or 8 south of east. (c) We use the result from part (a) in Eq. 4-8 along with the fact that Δt = 9 h. In unit ector notation, we obtain This leads to ag ag =.7 km/h. (6i ˆ ˆj) km = = (.67ˆi.ˆj) km/h. 9 h

8 (d) The direction of is θ = tan [(. km/h)/(.67 km/h)] = 8, or 8 south of east. ag (e) The aerage speed is distinguished from the magnitude of aerage elocity in that it depends on the total distance as opposed to the net displacement. Since the camel traels 4 km, we obtain (4 km)/(9 h) =.56 km/h.6 km/h. (f) The net displacement is required to be the 9 km East from A to B. The displacement from the resting place to B is denoted Δ r. 3 Thus, we must hae Δr + Δr + Δr = (9 km) ˆi 3 which produces Δ r 3 = (3 km)i ˆ+ ( km)j ˆ in unit-ector notation, or (36 33 ) in magnitude-angle notation. Therefore, using Eq. 4-8 we obtain (g) The direction of 36 km ag = =.km/h. ( 9) h ag is the same as r 3 (that is, 33 north of east).