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Cutnell Homework Solutions – Vectors 2D. Kinematics 12 – Homework Keys. PLO “C2” – Vector Kinematics in 2D (Constant Acceleration). Cutnell p.75 ( Chapter ...
Cutnell Homework Solutions – Vectors 2D

Kinematics 12 – Homework Keys PLO “C2” – Vector Kinematics in 2D (Constant Acceleration) Cutnell p.75 (Chapter Three) 12.

REASONING AND SOLUTION a. The x component of velocity is

vx = v0 x + a xt = 5480 m/s + (1.20 m/s 2 ) ( 842 s ) = 6490 m/s b. For the y component

v y = v0 y + a y t = 0 m/s + ( 8.40 m/s 2 ) ( 842 s ) = 7070 m/s 13.

REASONING AND SOLUTION Use the information concerning the x motion to find the time of flight of the ball x = voxt

or

t = x/vox = (19.6 m)/(28.0 m/s) = 0.700 s

The motion in the y direction is therefore subject to 2 2 2 y = voyt – (1/2)gt = 0 – (1/2)(9.80 m/s )(0.700 s) = –2.40 m The height of the tennis ball is

2.40 m .

14. REASONING The vertical component of the ball’s velocity v0 changes as the ball approaches the opposing player. It changes due to the acceleration of gravity. However, the horizontal component does not change, assuming that air resistance can be neglected. Hence, the horizontal component of the ball’s velocity when the opposing player fields the ball is the same as it was initially. SOLUTION Using trigonometry, we find that the horizontal component is

b

g

v x = v 0 cosθ = 15 m / s cos 55° = 8.6 m / s

Cutnell Homework Solutions – Vectors 2D

PLO “C” – Projectile Motion in 2D (Constant Acceleration) Cutnell p.75 (Chapter Three) 15.

SSM REASONING The time that the ball spends in the air is determined by its vertical motion. The time required for the ball to reach the lake can be found by solving Equation 3.5b for t. The motion of the golf ball is characterized by constant velocity in the x direction and accelerated motion (due to gravity) in the y direction. Thus, the x component of the velocity of the golf ball is constant, and the y component of the velocity at any time t can be found from Equation 3.3b. Once the x and y components of the velocity are known for a particular time t, the speed can be obtained from

v = v x2 + v y2 .

SOLUTION a. Since the ball rolls off the cliff horizontally, v0y = 0. If the origin is chosen at top of the cliff and upward is assumed to be the positive direction, then the vertical component of the ball's displacement is y = – 15.5 m. Thus, Equation 3.5b gives

t=

2y = ay

2( −15.5 m) = 1.78 s (–9.80 m / s 2 )

b. Since there is no acceleration in the x direction, vx

= v0 x = 11.4 m/s . The y component of the velocity of

the ball just before it strikes the water is, according to Equation 3.3b,

v y = v0 y + a y t = 0 + (–9.80 m/s 2 )(1.78 s)  = –17.4 m/s   The speed of the ball just before it strikes the water is, therefore,

v = v x2 + v y2 = (11.4 m / s) 2 + ( −17.4 m / s) 2 = 20.8 m / s 18.

REASONING a. The maximum possible distance that the ball can travel occurs when it is launched at an angle of 45.0°. When the ball lands on the green, it is at the same elevation as the tee, so the vertical component (or y component) of the ball's displacement is zero. The time of flight is given by the y variables, which are listed in the table below. We designate "up" as the +y direction. y-Direction Data y

ay

0m

2 −9.80 m/s

v0y

t

+(30.3 m/s) sin 45.0° = +21.4 m/s

?

vy

Since three of the five kinematic variables are known, we can employ one of the equations of kinematics to find the time t that the ball is in the air. b. The longest hole in one that the golfer can make is equal to the range R of the ball. This distance is given by the x variables and the time of flight, as determined in part (a). Once again, three variables are known, so an equation of kinematics can be used to find the range of the ball. The +x direction is taken to be from the tee to the green. x-Direction Data

Cutnell Homework Solutions – Vectors 2D

x

ax

R=?

2 0 m/s

v0x

t

+(30.3 m/s) cos 45.0° = +21.4 m/s

from part a

vx

SOLUTION a. We will use Equation 3.5b to find the time, since this equation involves the three known variables in the y direction:

(

)

y = v0 y t + 12 a y t 2 = v0 y + 12 a y t t 0 m =  +21.4 m/s +

1 2

( −9.80 m/s ) t  t 2

Solving this quadratic equation yields two solutions, t = 0 s and t = 4.37 s. The first solution represents the situation when the golf ball just begins its flight, so we discard this one. Therefore,

t = 4.37 s .

b. With the knowledge that t = 4.37 s and the values for ax and v0x (see the x-direction data table above), we can use Equation 3.5a to obtain the range R of the golf ball. 2

x = v0 x t + 12 a x t 2 = ( +21.4 m/s )( 4.37 s ) + 12 ( 0 m/s 2 ) ( 4.37 s ) = 93.5 m { =R

20.

REASONING AND SOLUTION The maximum vertical displacement y attained by a projectile is given by 2

Equation 3.6b ( v y

= v02y + 2a y y ) with vy = 0:

y=–

v02y 2a y

In order to use Equation 3.6b, we must first estimate his initial speed v0y . When Jordan has reached his maximum vertical displacement, vy = 0, and t = 1.00 s. Therefore, according to Equation 3.3b ( vy

= v0 y + a y t ), with upward taken as positive, we find that

v0 y = – a y t = – (–9.80 m/s 2 ) (1.00 s) = 9.80 m/s Therefore, Jordan's maximum jump height is

y=–

(9.80 m/s)2 2(–9.80 m/s 2 )

= 4.90 m

This result far exceeds Jordan’s maximum jump height, so the claim that he can remain in the air for two full seconds is false. 22.

REASONING AND SOLUTION The time required for the car to fall to the ground is given by

Cutnell Homework Solutions – Vectors 2D

t=

−2 y −2 ( −54 m ) = = 3.3 s g 9.80 m/s 2

2 During this time, the car traveled a horizontal distance of 130 m. Using ax = 0 m/s gives vox = x/t = (130 m)/(3.3 s) = 27.

39 m/s

REASONING AND SOLUTION The time of flight of the motorcycle is given by

2v0 sin θ 0

t=

g

=

2 ( 33.5 m/s ) sin18.0° = 2.11 s 9.80 m/s 2

The horizontal distance traveled by the motorcycle is then x = vo cos θo t = (33.5 m/s)(cos18.0°)(2.11 s) = 67.2 m The daredevil can jump over (67.2 m)/(2.74 m/bus) = 24.5 buses. In even numbers, this means 40.

24 buses .

REASONING Using the data given in the problem, we can find the maximum flight time t of the ball using Equation 3.5b (

1 2

y = v0 y t + a y t 2 ). Once the flight time is known, we can use the definition of average

velocity to find the minimum speed required to cover the distance x in that time. SOLUTION Equation 3.5b is quadratic in t and can be solved for t using the quadratic formula. According to Equation 3.5b, the maximum flight time is (with upward taken as the positive direction)

t=

– v0 y ± v02y – 4 2

=

() 1 2

( )a 1 2

y

(– y)

ay

– (15.0 m/s ) sin 50.0° ±

= 0.200 s

and

=

– v0 y ± v02y + 2a y y ay 2

(15.0 m/s ) sin 50.0°  +2(–9.80 m/s 2 ) (2.10 m) –9.80 m/s 2

where

the

2.145 s

first root corresponds to the time required for the ball to reach a vertical displacement of y = +2.10 m as it travels upward, and the second root corresponds to the time required for the ball to have a vertical displacement of y = +2.10 m as the ball travels upward and then downward. The desired flight time t is 2.145 s. During the 2.145 s, the horizontal distance traveled by the ball is

x = vxt = (v0 cos θ )t = [ (15.0 m/s) cos 50.0°] (2.145 s) = 20.68 m Thus, the opponent must move 20.68 m – 10.0 m = 10.68 m in 2.145 s – 0.30 s = 1.845 s . opponent must, therefore, move with a minimum average speed of

The

Cutnell Homework Solutions – Vectors 2D

v min =

46.

10.68 m = 5.79 m / s 1.845 s

REASONING AND SOLUTION Let H be the initial height of the can above the muzzle of the rifle. Relative to its initial position the vertical coordinate of the bullet at time, t, is

y = v0 y t − 12 gt 2 Relative to the can’s initial position the vertical coordinate of the can at the same time is 2 y' = (1/2)gt . NOTE: y = H – y' if the bullet hits the can. Then y = voyt – y'. The horizontal distance traveled by the bullet in time t is x = voxt. Solving for t and substituting gives y = (voy/vox)x – y' Now voy/vox = tan θ and it is seen from the figure that H = x tan θ if the bullet is to hit the can so y = H – y'. Hence, both objects are at the same place at the same time, and the bullet will always strike the can.