TUE, NOV 30^{th} PHYSICS (ACADEMIC)

1) Get out HW to be checked.
Go over answers.

2) Work on Projectile
Problems 2. Hand in at end of class or next class.

3) Read Lab Sheet on
Projectiles for next class. Remember that you will be working individually.

TEST TUE DEC 7^{th}!!!

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Projectile Problems 2:
Remember the steps and formulas…. Solve on a fresh piece of paper to leave
yourself room to work!!!

1)If a bullet from a gun is
shot horizontally at 500 mph (223.5 m/s), how far does it drop after 100
meters?

*Answer:Dy = - .98 m*

__X__

*Vx = 223.5 m/s*

*Dx=100 m*

*Dx=VxT*

*100 = 223.5 * T*

*T= 100/223.5 =.4474 sec*

* *

__Y__

*Viy = 0*

*Ay = -9.8 m/s ^{2}*

*T = .4474 sec*

*Dy = ViyT + ½ AyT ^{2}*

*Dy = 0(.4474) + ½
(-9.8)(.4474) ^{2}*

*Dy=-.98 m*

2) In the movie “The Gods
must be Crazy”, it begins with a pilot dropping a bottle out of an airplane. It
is recovered by a surprised native below, who thinks it is a message from the
gods. IF the plane from which the bottle was dropped was flying at an altitude
of 500 m, and the bottle lands 400m horizontally from the initial dropping
point, how fast was the plane flying when the bottle was released?

*Answer: Vx =39.6 m/s (gave
the wrong answer initially, sorry!!!)*

* *

__Y__

*Dy = - 500m*

*Ay = -9.8 m/s ^{2}*

*Viy = 0 m/s*

*T=??*

*Dy = ViyT + ½ AyT ^{2}*

*-500 = 0T+ ½ -9.8 T ^{2}*

*-500 = -4.9T ^{2}*

*-500/-4.9 = T ^{2}*

*T= 10.1 sec *

__X__

*Dx = 400 m*

*T = 10.1 sec*

*Dx=VxT*

*400 = Vx (10.1)*

*Vx = 400/10.1*

*Vx = 39.6 m/s*

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3) If I toss a marble into
the air at a velocity of 3.9 m/s at an angle of 50 degrees, and it reaches the
same height 0.6097 seconds later, how far did it travel horizontally?

* *

__X__

*Vx = V cos θ*

*Vx = 3.9 cos (50°)*

*Vx = 3.9 *.643*

*Vx= 2.5 m/s*

*Dx = Vx T*

*Dx =??*

*T = .6097*

*Dx = 2.5 (.6097)*

*Dx= 1.528 m*

* *

4) Jack be nimble, Jack be
quick, Jack jumped over the candlestick with a velocity of 5 m/s at an angle of
30 degrees. Did Jack burn his feet on the
0.25 m high candle?

* *

__Y__

*We want the distance at the top of the trip…. Dy = max height*

*Vfy=0*

*Ay = -9.8 m/s ^{2}*

*Viy = V sin θ*

*Viy = 5 sin(30°)*

*Viy =5 * .5*

*Viy = 2.5 m/s*

*Vfy ^{2}=Viy^{2} + 2 Ay Dy*

*0 ^{2} = 2.5^{2} + 2 (-9.8) Dy*

*-6.25 = -19.6 Dy*

*Dy = -6.25/-19.6*

*Dy =.319 m*

5) RANGE: How far away from a
target, level with you, do you have to stand to hit it, throwing a ball at 25
m/s at a 30 degree angle?

*Answer: Dx=55.23 m*

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__X__

*Vx = V cos θ*

*Vx = 25 cos 30°*

*Vx =21.65 m/s*

*Dx = Vx T*

*Dx = 21.65 T*

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__Y__

*Target is level, so Dy = 0*

*Ay = -9.8 m/s ^{2}*

*Viy = V sin θ*

*Viy = 25 sin 30°*

*Viy = 12.5 m/s*

*0 = 12.5 T + ½(-9.8)T ^{2}*

*0 = T (12.5 + -4.9 T)*

*0 = 12.5 + -4.9 T*

*-12.5 = -4.9T*

*T = -12.5/-4.9*

*T =2.551 sec*

*Back to X*

* *

*Dx = 21.65 T*

*Dx = 21.65 (2.551)*

*Dx = 55.23 m*

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6) Minnie jumps up off a 14
meter high cliff at a 20 degree angle with a speed of 5 m/s. Mickey is at the bottom
of the cliff, 55 meters away from the bottom. He starts running when she jumps
. How long does it take Minnie to reach the bottom? How fast should Mickey run
to catch her?

*Answer: T=1.87 sec,
Vmickey=29.411 m/s ( to the cliff, or Vx = 24.65 m/s to exactly catch her)*

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__Y for Minnie__

*Dy = - 14 m*

*Ay = -9.8 m/s ^{2}*

*Viy = V sin θ*

*Viy = 5 sin 20°*

*Viy = 1.71 m/s*

*Dy = ViyT + ½ AyT ^{2}*

*-14 = 1.71 T + ½(-9.8)T ^{2}*

*0 = (-4.9)T ^{2} +
1.71T + 14*

*Using quadratic formula or
math solver*

*T =1.874 sec*

__X for Minnie__

*Vx = V cos θ*

*Vx = 5 cos 20°*

*Vx =4.698 m/s*

*Dx = Vx T*

*Dx = 4.698 (1.874)*

*Dx = 8.8 m*

* *

__Mickey in the X__

*Dx = 55 – 8.8 = 46.2
m*

*Dx = Vx T*

*46.2 = Vx (1.874)*

*Vx = 24. 65 m/s*

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