* *

* *

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

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}!!!

---------------------------------

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? How about shooting at an angle of 45 degrees?

*Answer:Dy = - .98 m, 98.03
m (up)*

*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*

--------------------------------------------

* *

* *

__X__

*Vx = V*

*Vx = 223.5 cos (45°)*

*Vx = 223.5 *.707*

*Vx= 158 m/s*

* *

*Dx=100 m*

*Dx=VxT*

*100 = 158 * T*

*T= 100/223.5 =.6329 sec*

* *

__Y__

*Viy = V sin θ*

*Viy = 223.5 sin(45°)*

*Viy =223.5 * .707*

*Viy = 158.03 m/s*

* *

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

*T = .6329 sec*

*Dy = ViyT + ½ AyT ^{2}*

*Dy = 158.03(.6329) + ½
(-9.8)(.6329) ^{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*

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 some seconds later, how far did it travel horizontally?

*Answer: Dx=1.528 m*

* *

__Y__

*Vy = V sin θ*

*Vy = 3.9 sin (50°)*

*Vy = 3.9 *.766*

*Viy= 2.98757 m/s*

*Dy=0*

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

*Dy = ViyT + ½ AyT ^{2}*

*0 = 2.988 T + ½(-9.8)T ^{2}*

*0 = T (2.988 + -4.9 T)*

*0 = 2.988 + -4.9 T*

*-2.988 = -4.9T*

*T = -2.988/-4.9*

*T =0.6097 sec*

*Back to X*

* *

__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?

*Answer: Dy = .319 m, no!*

* *

* *

__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, 2 m above you, do you have to stand to hit it, throwing a ball at 25 m/s
at a 30 degree angle?

*Answer: Dx=3.7m or 51.5 m*

__Y __

*Dy = - 2 m*

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

*Viy = V sin θ*

*Viy = 25 sin 30°*

*Viy = 12.5 m/s*

*Dy = ViyT + ½ AyT ^{2}*

*-2 = 12.5 T + ½(-9.8)T ^{2}*

*0 = (-4.9)T ^{2} +
12.5T + 2*

*Using quadratic formula or
math solver*

*T =.171 sec or 2.38 sec
(two times when it is at the height, once on the way up, and once on the way
down!)*

__X __

*Vx = V cos θ*

*Vx = 25 cos 30°*

*Vx =21.65 m/s*

*Dx = Vx T*

*Dx = 21.65 (.171) or
Dx=21.65(2.38)*

*Dx = 3.7 m or 51.527m*

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
reaches the apex of her jump. How fast should he 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)*

* *

* *

* *

__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*

* *

THINKING QUESTION:

A
monkey is in a 10 m high tree and falls. A hunter is on the ground 20 m away
aiming directly at the monkey. How fast does he have to fire the gun to hit the
monkey as it falls? Draw a picture and explain. USE ALGEBRA TO PROVE YOUR
ANSWER!

Hunter:

*Θ = tan ^{-1}
(TreeHeight/Dx) = tan^{-1} (10/20) =26.565°*

*Vix= V*

*Dx = Vx T*

*Dx = 20m*

*20 = V*

*T = 20 /( V*

* *

*Dy (bullet) =TreeHeight +
Dy monkey*

*Viy (bullet) = V*

*Bullet:*

*Dy = Viy T + ½ Ay T ^{2}*

*Dy = V*

*Dy = V*

* *

*Dy monkey = ½ Ay T ^{2}
=- 4.9 (20 /( V*

* *

*10 +- 4.9 (20 /( V*

*simplifies*

*10 = V*

*10 = V/V * *

*10 = 10 !!!!!*

*Every Velocity is
correct!!!*

* *

* *

* *

* *

* *

* *

* *

* *

*The monkey and bullet are
both falling at the same rate… if you looked at them in the reference frame of
the acceleration of gravity, as long as the bullet was pointed at the monkey to
begin with, it would always hit it no matter what the speed of the bullet!*

* *

* *

*ALGBEBRA PROOF*

*Θ = tan ^{-1}
(TreeHeight/Dx) = tan^{-1} (H/Dx) *

*Vix= V*

*Dx = Vx T*

*Dx = V*

*T =( Dx /( V*

* *

*Dy (bullet) =TreeHeight +
Dy monkey*

*Viy (bullet) = V*

* *

*Bullet:*

*Dy = Viy T + ½ Ay T ^{2}*

*Dy = V*

* *

*Dy = V*

* *

*Dy monkey = ½ Ay T ^{2}
=- 4.9 (( Dx /( V*

* *

* H +- 4.9 (( Dx /( V*

^{ }

*Simplifies*

* *

*H == V*

* *

*H = V/V **

* *

*H =1 * Dx * tan (tan-1
(H/Dx) )*

* *

*H= Dx *H/Dx*

* *

*H=H!!!*

* *

If you are standing 5 m away
from a target that is 2 meters above you, what angle do you need to hold a gun
that shoots 25 m/s to hit it on the way down?

*Θ=??*

__X__

* *

*Dx = 5 m*

*Vx = V*

*Vx = 25 cos θ*

*Dx = Vx T*

*5 = 25
cos θ T*

*T = 5/(25 cos θ)*

*T = .2/ cos θ*

* *

__Y__

*Dy = 2 m*

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

*Viy = V *

*Viy = 25 sin θ*

* *

*Dy = Viy T
+ ½ Ay T ^{2}*

* *

*2 = 25 sin θ (.2/ cos θ) + ½ (-9.8) (.2/ cos θ) ^{2}*

* *

*0 = 5 sin θ /cos
θ + -4.9 (.04) * (1/cos θ) ^{2}
-2*

*use math solver??? Θ
= 24.09° or 87.7°*

*Or keep going and remember
from crazy trig identities that*

* sin/cos = tan and 1/cos ^{2} = sec^{2} = 1
+tan^{2}*

* *

*0 = 5 tan(θ) + 0.196 ( 1 + tan ^{2}(θ) )
-2*

*0 = -0.196 (tan(θ)) ^{2}
+ 5 (tan (θ)) - 1.804*

*quadratic formula for tan
(θ) = .447034 or 25.06*

*so θ = tan ^{-1}
(.44703) = 24.09°*

* or θ= tan ^{-1}(25.06) = 87.7°*

*correct answer: 87.7° to
hit it on the way down.*