10 extra topic: Projectile Motion
Greg Kelly, Hanford High School, Richland, Washington
Photo by Vickie Kelly, 2002
Fort Pulaski, GA
One early use of calculus was to study projectile motion.
In this section we assume ideal projectile motion:
Constant force of gravity in a downward direction
Flat surface
No air resistance (usually)
We assume that the projectile is launched from the origin at time t =0 with initial velocity vo.
ov
Let o ov v
then cos sino o ov v v i j
The initial position is: r 0 0 0o i j
ov
Newton’s second law of motion:
Vertical acceleration
f ma2
2f
d rm
dt
ov
Newton’s second law of motion:
The force of gravity is:
Force is in the downward direction
f ma f mg j2
2f
d rm
dt
ov
Newton’s second law of motion:
The force of gravity is:
f ma f mg j2
2f
d rm
dt
mg j2
2
d rm
dt
ov
Newton’s second law of motion:
The force of gravity is:
f ma f mg j2
2f
d rm
dt
mg j2
2
d rm
dt
2
2
d rg
dt j
Initial conditions: r r v when o o
drt o
dt
2o
1r v r
2 ogt t j
21r
2gt j 0 cos sin o ov t v t i j
o vdr
gtdt
j
21r cos sin 0
2 o ogt v t v t j i j
21r cos sin
2o ov t v t gt
i j
Vector equation for ideal projectile motion:
21r cos sin 0
2 o ogt v t v t j i j
21r cos sin
2o ov t v t gt
i j
Vector equation for ideal projectile motion:
Parametric equations for ideal projectile motion:
21cos sin
2o ox v t y v t gt
Example 1:A projectile is fired at 60o and 500 m/sec.Where will it be 10 seconds later?
500 cos 60 10x
2500x
21500sin 60 10 9.8 10
2y
3840.13y
Note: The speed of sound is 331.29 meters/secOr 741.1 miles/hr at sea level.
The projectile will be 2.5 kilometers downrange and at an altitude of 3.84 kilometers.
The maximum height of a projectile occurs when the vertical velocity equals zero.
sin 0o
dyv gt
dt
sinov gt
sinovt
g
time at maximum height
The maximum height of a projectile occurs when the vertical velocity equals zero.
sin 0o
dyv gt
dt
sinov gt
sinovt
g
We can substitute this expression into the formula for height to get the maximum height.
2
max
sin sin1sin
2o o
o
v vy v g
g g
21sin
2oy v t gt
2 2
max
sin sin
2o ov v
yg g
2
max
sin sin1sin
2o o
o
v vy v g
g g
21sin
2oy v t gt
2 2
max
si2
2
n sin
2o ov v
yg g
2
max
sin
2ov
yg
maximum
height
210 sin
2ov t gt When the height is zero:
10 sin
2ot v gt
0t time at launch:
210 sin
2ov t gt When the height is zero:
10 sin
2ot v gt
0t time at launch:1
sin 02ov gt
1sin
2ov gt
2 sinovt
g
time at impact
(flight time)
If we take the expression for flight time and substitute it into the equation for x, we can find the range.
cos ox v t
cos 2 sin
oov
x vg
If we take the expression for flight time and substitute it into the equation for x, we can find the range.
cos ox v t
2 sincos o
o
vx v
g
2
2cos sinovx
g
2
sin 2ovx
g Range
The range is maximum when
2
sin 2ovx
g Range
sin 2 is maximum.
sin 2 1
2 90o
45o
Range is maximum when the launch angle is 45o.
If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.
cosox v t
coso
xt
v
21sin
2oy v t gt
If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.
cosox v t
coso
xt
v
21sin
2oy v t gt
2
1sin
2cos cosoo o
x x
vy g
vv
This simplifies to:
22 2
tan2 coso
gy x x
v
which is the equation of a parabola.
If we start somewhere besides the origin, the equations become:
coso ox x v t 21sin
2o oy y v t gt
Example 4:A baseball is hit from 3 feet above the ground with an initial velocity of 152 ft/sec at an angle of 20o from the horizontal. A gust of wind adds a component of -8.8 ft/sec in the horizontal direction to the initial velocity.
The parametric equations become:
152cos 20 8.8ox t 23 152sin 20 16oy t t
ov oy 1
2g
These equations can be graphed on the TI-89 to model the path of the ball:
Note that the calculator is in degrees.
t2
Max height about 45 ft
Distance traveled about 442 ft
Timeabout
3.3 sec
Usingthe
tracefunction:
In real life, there are other forces on the object. The most obvious is air resistance.
If the drag due to air resistance is proportional to the velocity:
dragF kv (Drag is in the opposite direction as velocity.)
Equations for the motion of a projectile with linear drag force are given on page 546.
You are not responsible for memorizing these formulas.
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