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Unformatted text preview: Chapter 3 Homework Solutions P: 1, 7, 9, 11, 13 P1. A car is driven 215 km west and then 85 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram. Solution:
N W D2y D2x 45 D1 =215 km D2 =85 km D E North S Choose: North is "up" and "+" East is "+" km km cos sin (points west) km (points west) km (points south) km km km tan
km km km km km km km (pointing south of west) P7. V is a vector 14.3 units in magnitude and points at an angle of 34.8 above the negative x axis. (a) Sketch this vector. (b) Find V and V (c) Use V and V to obtain (again) the magnitude and direction of V 1 Solution: (a)
y 14.3 34.8 Vx Vy x (b) cos sin (c) units tan sin cos units units P9. An airplane is traveling 735 km/h in a direction 41.5 west of north. (a) Find the components of the velocity vector in the northerly and westerly directions. (b) How far north and how far west has the plane traveled after 3.00 h? Solution:
N y 735 km/h W 41.5 Vy E x Vx S (a) sin sin km/h sin km h 2 cos cos (b) , km km km/h cos
km h km h ! " 3h " 3h km h km km y B=26. 5 Ay 56.0 Bx 28. 0 Ax x By A=44. 0 C=31. 0 P11. Determine the vector A Solution:
y B=26. 5 Ay 56.0 Bx By C , given the vectors A and C in the Fig. 1 A=44. 0 28. 0 Ax x C=31. 0 A C is drawn in red on the diagram. It is a vector that points from the tip of C to the tip of A We can determine its magnitude and direction by adding the components of A and C : # cos # cos # # sin # sin # 3 $ $ # # $ $ # # $ $ The magnitude of the vector then is: % A $ % The direction is determined by the angle between the vector and the x axis: tan
# $ # $ P13. For the vectors given in the Fig. 1, determine (a) # & $ '()# & $ ' )$ # & Solution: We first calculate the components of the three vectors: # # & & $ $ (a) # # # % # & & & & $ $ $ $ % ? # # & & $ $ cos sin sin cos tan Note:Alternative answer for the angle: The ** ** in the angle means that the vector is below the x-axis. is equavalent to 4 (b) # # # % # tan & & & & $ $ $ $ % ? # # & & $ $ ? $ # & $ ( ) # & $ Notice that # & this vector is going to have the same magnitude but opposite direction to the vector from (b). Therefore: # # % # & & & $ $ $ % # # & & $ $ tan Notice that the angle turns out to be the same as (b). The x and y components, however, are negative meaning that the vector lies in III quadrant, while the vector in (b) has all positive components and lies in I quadrant. Therefore, the actual angle measured from the positive direction of the x axis is: 5 P: 19, 21, 31, 33, 37*, 39* Example 3-8 : Level horizontal range (a) Derive a formula for the horizontal range R of a projectile in terms of its initial velocity ! and angle . The range means what is the horizontal distance traveled before the projectile returns to its initial height. (b) At what angle should it have been aimed to strike a target 320 m away. Solution: (a)
y o=60 o=45 o=30 x Range Choose: 1. s 2. m 3. m 4. "up" is "+" + m/s Given: 1. ! initial velocity initial angle Find: R=? What is the range (how far does the water go?) Solution: Along y: + ! + m m ! + ! m s (initial time) + ! m ! (the time it takes for the total flight) + 6 Along x: ! ,
! ! ! ! m/s sin cos ! ! !
sin cos ! + ! ! + ! sin ! cos
sin Note: We have used the trig. relation : sin cos sin P19. A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s) should the nozzle point in order that the water land 2.0 m away? Why are there two different angles? Solution: We can use the formula for the range from above (it is also derived in the Textbook, Example 3-8). Given: Range , m ! m/s Find: ? Solution: ! sin , + m sin sin Note: cos sin sin sin is also a solution
m/s sin m/s m m/s m/s sin cos sin sin cos cos The angle Thus we have two solutions: 7 P21. A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the base. What was the ball's initial speed? Solution: Choose: 1. s 2. m 3. m 4. "up" is "+" + m/s Given: the time when it falls on the ground ') m (distance from the building ) 3. ' ) m (ground) 4. ! m/s (initial velocity has no y component) Find: ! ? Solution: Along y: The y component of the final velocity is: ! ! + ! m/s m/s m !
m s m m/s The time it took to fall on the ground then is: ! ! + m/s m/s m/s s We can find the initial velocity then from the position along x: ! m m ! s ! ! m/s P31. A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37.0 with the horizontal, as shown on the diagram. (a) Determine the time taken by the projectile to hit point P at ground level (b) Determine the range X of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point P, find (c) the horizontal and the vertical component of its velocity (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal (f) Find the maximum height above the cliff top reached by the projectile Solution: 8 y v o=65.0 m/s 37.0 125 m x Choose: s m m "up" is positive + m/s cos sin m/s Given: !
! ! ! ! ! ! m/s cos m/s sin m/s m/s (a) ? for the projectile to hit the ground ! m m/s
m/s . m/s . -9.8 m/s + m m/s m/s
m/s m/s m/s m/s m/s m
m m 255 s (the negative value has no physical meaning) (b) Determine the range X of the projectile as measured from the base of the cliff. ! 9 / m m/s 255 s (c) ! -! ? just before it hits the ground ! ! m/s ! ! + m/s m/s (d) ! ? just before it hits the ground ! (e) ! ! m/s 41.084 m 255 s 3 01 m/s 3.07 m/s 679 m/s ? just before it hits the ground ! 3.07 m/s tan ! 51.9 m/s The angle is negative because the velocity is pointing downward. (f) 0 ? ! ! +0 ! +0 0 0
! + m/s m/s m P33. At what projection angle will the range of a projectile equal its maximum height? Solution: ,
! sin + ! + 'see Example 3-8) 0 (see P31, part f) In order for the range to be equal to the max. height, we need: , 0 remember that sin
! sin + cos sin ! sin + ! + ! sin + sin cos ! sin + cos tan 10 ...
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- Fall '09