NAME_______                      ANSWERS BOOK WORK CHAPTER 3 Projectiles

 

Pg. 102

 

1) A pelican flying along a horizontal path drops a fish from a height of 5.4 m while traveling 5 m/s. How far does the fish travel horizontally before it hits the water below?

 

X

Vx = 5 m/s

Dx =??

T=??

Dx=Vx T

 

Y

Viy = 0 m/s

Ay = -9.8 m/s2

Dy= -5.4 m

 

Dy = Viy T + Ay T2

-5.4 = 0 (T) + (-9.8) T2

-5.4 = -4.9 T2

T= 1.05 sec

Back to X

Dx=5(1.05)=5.25 m

 

 

 

 

 

 

 

 

 

 

2) Give both the horizontal and vertical components of the velocity of the fish from item 1 before the fish enters the water. ( Find Vfx and Vfy if Dy= -5.4m, Vix= 5 m/s , Viy=0 )

X

Vfx=Vx= 5 m/s

Y

Dy = -5.4 m

Ay = -9.8 m/s2

Viy = 0 m/s

Vfy = ????

 

Vfy2=Viy2 + 2 Ay Dy

Vfy2=(0) + 2 (-9.8) ( -5.4)

Vfy2=105.84

Vfy = - 10.29 m/s (downward)

 

For the final velocity

V2 = Vx2+ Vy2

V2 = 52 + (10.29)2

V= 11.44 m/s  θ = tan -1 ( Vy/Vx) =

tan -1 (10.29/5) = 64 degrees to the water

 

Final Velocity is 11.44 m/s at 64 degrees to the water

 

 

 

 

 

 

 

 

 

 

 

 

3) Find the instantaneous velocity of the stunt dummy in Sample Problem 3D as it hits the water. ( Find Vfx and Vfy if Dy= - 10 m, Vix= 22.5 m/s , Viy=0 )

X

Vfx=Vx= 22.5 m/s

Y

Dy = -10 m

Ay = -9.8 m/s2

Viy = 0 m/s

Vfy = ????

 

Vfy2=Viy2 + 2 Ay Dy

Vfy2=(0) + 2 (-9.8) ( -10)

Vfy2=196

Vfy = - 14 m/s (downward)

 

For the final velocity

V2 = Vx2+ Vy2

V2 = 22.52 + (14)2

V= 26.5m/s    

θ = tan -1 ( Vy/Vx) = tan -1 (14/22.55) = 31.9 degrees to the ground

Final Velocity is 25.5 m/s at 31.9 degrees to the ground

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4) A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way and the cat slides off the table at a speed of 5 m/s. Where does the cat strike the floor?

X

Vx = 5 m/s

Dx =??

T=??

Dx=Vx T

 

Y

Viy = 0 m/s

Ay = -9.8 m/s2

Dy= -1 m

 

Dy = Viy T + Ay T2

-1 = 0 (T) + (-9.8) T2

-1 = -4.9 T2

T= .45 sec

Back to X

Dx=5(.45)=2.26 m

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ANSWERS Bookwork Chapter 3 pg. 105 Section Review (do on fresh paper so you have room!)

 

1) Which of the following are examples of projectile motion?

a) airplane taking off, b) tennis ball lobbed over a net. c) plastic disk sailing over the lawn d) a hawk diving to catch a mouse e) a parachutist drifting to Earth f) a frog jumping from the land into the water.

 

2) Which of the following exhibit parabolic motion?

a) a flat rock skipping over the surface of the lake b) a three point shot in basketball c) the space shuttle while orbiting the Earth d) a ball bouncing across the room e) a cliff diver f) a life preserver dropped from a stationary helicopter g) a person skipping

 

3) An Alaskan rescue plane drops a package of emergency rations to s stranded party of explorers. The plane is traveling horizontally at 100 m/s at a height of 50 m above the ground. What horizontal distance does the package travel before striking the ground?

X

Vx = 100 m/s

Dx =??

T=??

Dx=Vx T

 

Y

Viy = 0 m/s

Ay = -9.8 m/s2

Dy= -50 m

 

Dy = Viy T + Ay T2

-50 = 0 (T) + (-9.8) T2

-50 = -4.9 T2

T= 3.19 sec

Back to X

Dx=100 (3.19)=319 m

 

 

4) Find the velocity (vector magnitude and direction) of the package in item 3 just before it hits the ground.

X

Vfx=Vx= 100 m/s

Y

Dy = -50 m

Ay = -9.8 m/s2

Viy = 0 m/s

Vfy = ????

 

Vfy2=Viy2 + 2 Ay Dy

Vfy2=(0) + 2 (-9.8) ( -50)

Vfy2= 980

Vfy = - 31.3 m/s (downward)

 

For the final velocity

V2 = Vx2+ Vy2

V2 = 1002 + (31.3)2

V= 104.7 m/s 

θ = tan -1 ( Vy/Vx) = tan -1 (31.3/100) = 31.9 degrees to the ground

Final Velocity is 104.7  m/s at 17.38 degrees to the ground

 

5) During a thunderstorm, a tornado lifts a car to a height of 125 m above the ground. Increasing in strength, the tornado flings the car horizontally with an initial speed of 90 m/s. How long does the car take to reach the ground? How far horizontally does the car travel before hitting the ground?

X

Vx = 90 m/s

Dx =??

T=??

Dx=Vx T

 

Y

Viy = 0 m/s

Ay = -9.8 m/s2

Dy= -125 m

 

Dy = Viy T + Ay T2

-125 = 0 (T) + (-9.8) T2

-125 = -4.9 T2

T= 5.05 sec

Back to X

Dx=90 ( 5.05)=454.57 m

 

 

 

 

 

 

 

 

 

 

 

*** 6) Streams of water in a fountain shoot from one level to the next. A particle of water in a stream takes 0.5 seconds to travel between the first and the second level. The receptacle on the second level is a horizontal distance of 1.5 m away from the spout on the first level. If the water is projected at a 33 angle, what is the initial speed of the particle?

X

Dx = 1.5 m

T = 0.5 sec

Dx = Vx T

1.5 = Vx (.5)

Vx = 3 m/s

Vx = V cos (θ)

3 = V  cos (33)

3 = V (.83867)

V =  3.577 m/s

 

 

 

 

 

 

 

 

 

 

 

 

*** 7) If a water particle in a stream of water in a fountain takes 0.35 seconds to travel from spout to receptacle when shot at an angle of 67 and an initial speed of 5 m/s, what is the vertical distance between the levels of the fountain?

Y

 

Vy = V sin (θ)

Vy = 5 sin (67)

Viy = 4.6 m/s

Ay = - 9.8 m/s2

T = 0.35 sec

Dy = ????

Dy = Viy T + Ay T2

 Dy = 4.6 (.35) + (-9.8) (.35)2

Dy = 1.61 + -4.9 *.1225

Dy = 1.00975 m