10.4 Projectile Motion
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
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:
21r cos sin
2o ov t v t gt
i j
Vector equation for ideal projectile motion:
Vector equation for ideal projectile motion:
Parametric equations for ideal projectile motion:
Example 1:A projectile is fired at 60o and 500 m/sec.Where will it be 10 seconds later?
Note: The speed of sound is 331.29 meters/secOr 741.1 miles/hr at sea level.
The maximum height of a projectile occurs when the vertical velocity equals zero.
The maximum height of a projectile occurs when the vertical velocity equals zero.
We can substitute this expression into the formula for height to get the maximum height.
21sin
2oy v t gt
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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