Projectiles Fired at an Angle by 1Le6cE

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									             Projectiles Fired at an Angle
Now let’s find range of a projectile fired with speed v0 at an angle .
    Step 1: Split the initial velocity vector into components.


                                v0
     vy = v0 sin
                            

                             vx = v0 cos


               And v0 = vx +      vy
             Projectiles Fired at an Angle                (cont.)
                Step 2: Find hang time.
ALL THINGS
 VERTICAL       Use y = v0t + ½ a t 2 with only vertical data:
                     y = (v0 sin) t + ½ (-a)t 2
             Over level ground, y = 0.
             Divide through by t: 0 = v0 sinθ - ½(9.8m/s 2) t,
             Rearrange equation: t = (v0 sinθ) / 4.9 m/s 2

                                            Note: If we had shot the
                             v0           projectile from a 100 m cliff,
                                              y would be -100 m.
    v0 sin
                         θ

                             v0 cos
       Projectiles Fired at an Angle                  (cont.)

ALL THINGS HORIZONTAL
Step 3: Now that we know how long it’s in the air, we know how
long it travels horizontally. (The projectile’s vertical and horizontal
movements are completely independent.)
Use Δx = v0t + ½ a t 2 again, this time with only horizontal data:

                         Δx = (v0 cos) t + ½ (0) t 2
                             = (v0 cos) t
            v0                  This is the same as saying:
                        horiz. distance = horiz. speed  time
        θ                         In other words, d = v t
            v0 cos
                  Symmetry and Velocity
- The projectile’s speed is the same at points directly across the parabola
(at the same vertical position). The angle is the same too, but with
opposite orientation.
                                         - Horizontal
                                         speeds are the
                                         same throughout
                                        the trajectory.
                
                                                        - Vertical speeds
                                                      are the same only at
            The vert. comp. shrinks then
          grows in opposite direction at a                points of equal
           const. rate (-g). The resultant                    height.
           velocity vector’s orientation
          and magnitude changes, but is
                 always tangent.
  θ
                     Symmetry and Time
Over level ground, the time at the peak is half the hang time. Notice the
symmetry of times at equal heights relative to the 10 unit mark. The
projectile has covered half its range when it has peaked, but only over
level ground.
                                  t = 10

                         t=5               t = 15


                   t=3                              t = 17




           t=0                                               t = 20
    Max height & hang time depend only on
            initial vertical velocity
- Each initial velocity vector below has the a different magnitude
(speed) but each object will spend the same time in the air and reach
the same max height.
- This is because each vector has the same vertical component.

-The projectiles will have different ranges, however. The greater the
horizontal component of initial velocity, the greater the range.
                           Max Range
- Over level ground at a constant launch speed, what angle maximizes
the range, R ?
- First consider some extremes: When θ = 0, R = 0, since the object is
on the ground from the moment it’s launched.
- When θ = 90, the object goes straight up and lands right on the launch
site, so R = 0 again.
         The best angle is 45, smack dab between the extremes.
                                         Here all launch speeds are the
                                         same; only the angle varies.
   76
           45

              38
           Range Formula & Max Range at 45
First find the time.
- Note that Δy = 0, since the projectile starts and stops at ground level
(no change). Δy = v0 t + ½ at 2.
- So, 0 = (v0 sinθ) t - ½ a t 2
- We divide through by t giving us t = 2 v0 sinθ/ a. Then,
                    R = (v0 cosθ) t
                      = (v0 cosθ ) (2 v0 sinθ / a)
                      = (2 v02 sinθcosθ) / a.

By the trig identity sin 2θ = 2 sinθcosθ, we get R = (v02 sin 2θ) / a.
Since v0 and a are fixed, R is at a max when sin 2θ is at a max.
When the angle, 2θ, is 90, the sine function is at its maximum of 1.
Therefore, θ = 45.
http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/Proj
ectileMotion/jarapplet.html

http://www.walter-fendt.de/ph11e/projectile.htm

C:\Documents and Settings\hstdilsaver\Local Settings\Temp\phet-
projectile-motion\projectile-motion_en.html

http://phet.colorado.edu/simulations/sims.php?sim=projectile_motion




                                                         9
           Max Range when Δy=0
- When fired from a cliff, or from below ground, a
projectile doesn’t attain its max range at 45.
- 45 is only the best angle when a projectile is fired over
level ground.
- When fired from a cliff, a projectile attains max range
with a launch angle less than 45  (see next slide).
- When fired from below ground, a projectile attains max
range with a launch angle greater than 45 .
       Range when fired from cliff
45                                       If ground were
                                           level, the 45
                                        launch would win.
        < 45

      Launch speeds
       are the same.

                Because the < 45
               parabola is flatter it
              eventually overtakes
                 45  parabola.
Ranges at complementary launch angles
          - An object fired at angle θ will have the same
   75
          range as when it’s fired at the same speed at an
          angle 90 - θ.
             Reason: R = 2v02 sinθ cosθ / a, and the sine
             of an angle is the cosine of its complement (and
             vice versa).
    15
                For example, R at 40 is
                = 2v02 sin 40 cos 40 / a
                 = 2v02 sin ( 90 - 40) cos ( 90 - 40)/a
                 = 2v02 cos 50 sin 50 / a
                        = R at 50.
 50

   40
                   Monkey in a Tree
Here’s a classic physics problem: You want to shoot a
banana at a monkey up in a tree. Knowing that the monkey
will get scared and let go of the branch the instant he hears
the sound of the banana gun, how should you aim: a little
above, a little below, or right at him?



               Monkey in a tree web site
          Monkey in a tree explanation
The reason you should aim right at the monkey even though the
monkey lets go right when you pull the trigger is because both the
monkey and the banana are in the air for the same amount of time
before the collision. So, with respect to where they would have
been with no gravity, they fall the same distance.

                                “gravity-free” monkey




                                              monkey w/ gravity
                       Homerun Example
From home plate to the center field wall at a ball park is 130 m. When a
batter hits a long drive the ball leaves his bat 1 m off the ground with a
velocity of 40 m/s at 28 above the horizontal. The center field wall is
2.6 m high. Does he hit a homerun?


      28                                                         2.6 m
          }   1m
                                   130 m
 Let’s first check the range to see if it even has a chance:
 R = v02 sin 2θ/ a
  = 402 sin 56° / 9.8 m/s2
   = 135.35 m.
We need to determine its vertical position when its horizontal position is
130 m. If it’s 1.6 m or more, it’s a homer. Let’s first find the time when
the ball is 130 m away (horizontally) from the point where it was hit.
                      Homerun       (cont.)

 t = d/v
   = (130 m) / (40 cos 28)
   = 3.68085 s.
Let’s see how high up it is at this time:
  Δy = v0sinθt + ½ at   2.

  Δy = (40 sin 28) (3.68085) - 4.9 (3.68085)2
     = 2.73 m
     which is 3.73 m above the ground, out of the reach
     of a leaping outfielder. Therefore, it’s a home run!

								
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