Projectiles Fired at an Angle

1
Projectiles Fired at an Angle
Now lets find range of a projectile fired with
speed v0 at an angle ?.
Step 1 Split the initial velocity vector into
components.
And v0 vx vy
2
Projectiles Fired at an Angle (cont.)

Step 2 Find hang time.
Use ?y v0t ½ a t 2 with only vertical data
? y (v0 sin?) t ½ (-a)t 2
ALL THINGS VERTICAL
Over level ground, ?y 0. Divide through by t
0 v0 sin? - ½(9.8m/s 2) t, Rearrange equation
t (v0 sin?) / 4.9 m/s 2
Note If we had shot the projectile from a 100 m
cliff, ? y would be -100 m.
3
Projectiles Fired at an Angle (cont.)

ALL THINGS HORIZONTAL
Step 3 Now that we know how long its in the
air, we know how long it travels horizontally.
(The projectiles vertical and horizontal
movements are completely independent.)
Use ?x v0t ½ a t 2 again, this time with
only horizontal data
?x (v0 cos?) t ½ (0) t 2 (v0 cos?) t
v0
This is the same as sayinghoriz. distance
horiz. speed time In other words, d v t
v0 sin?
?
v0 cos?
4
Symmetry and Velocity
- The projectiles 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 points of
equal height.
?
The vert. comp. shrinks then grows in opposite
direction at a const. rate (-g). The resultant
velocity vectors orientation and magnitude
changes, but is always tangent.
?
5
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.
6
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.
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Max Range
- Over level ground at a constant launch speed,
what angle maximizes the range, R ?
- First consider some extremes When ? 0, R
0, since the object is on the ground from the
moment its launched.
- When ? 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.
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Range Formula Max Range at 45
First find the time. - Note that ?y 0, since
the projectile starts and stops at ground level
(no change). ?y v0 t ½ at 2. - So, 0 (v0
sin?) t - ½ a t 2 - We divide through by t
giving us t 2 v0 sin?/ a. Then,
R (v0 cos?) t
(v0 cos? ) (2 v0 sin? / a)
(2 v02 sin?cos?) / a.
By the trig identity sin 2? 2 sin?cos?, we get
R (v02 sin 2?) / a. Since v0 and a are
fixed, R is at a max when sin 2? is at a max.
When the angle, 2?, is 90, the sine function is
at its maximum of 1. Therefore, ? 45.
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http//galileoandeinstein.physics.virginia.edu/mor
e_stuff/Applets/ProjectileMotion/jarapplet.html h
ttp//www.walter-fendt.de/ph11e/projectile.htm C
\Documents and Settings\hstdilsaver\Local
Settings\Temp\phet-projectile-motion\projectile-mo
tion_en.html http//phet.colorado.edu/simulations
/sims.php?simprojectile_motion
10
Max Range when ?y0
- When fired from a cliff, or from below ground,
a projectile doesnt 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 .
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Range when fired from cliff
12
Ranges at complementary launch angles
- An object fired at angle ? will have the same
range as when its fired at the same speed at an
angle 90 - ?.
Reason R 2v02 sin? cos? / a, and the sine of
an angle is the cosine of its complement (and
vice versa).
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.
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Monkey in a Tree
Heres 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
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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
banana path without gravity
monkey w/ gravity
path with gravity
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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?
40 m/s
2.6 m
28
1 m

Lets first check the range to see if it even has
a chanceR v02 sin 2?/ 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 its 1.6 m
or more, its a homer. Lets first find the time
when the ball is 130 m away (horizontally) from
the point where it was hit.
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Homerun (cont.)
t d / v
(130 m) / (40 cos 28)
3.68085 s.
Lets see how high up it is at this time
?y v0sin?t ½ at 2.
?y (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,
its a home run!