10.4 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 originat time t =0 with initial velocity vo.
The initial position is:
Horizontal component
Vertical component
At the origin
Note: Wewill seethis again
Newton’s second lawof motion:
Vertical acceleration
Force = Mass (Acceleration)
Second derivative = acceleration
Newton’s second lawof motion:
The force of gravity is:
Force is negative
Because gravity pulls downward
Newton’s second lawof motion:
The force of gravity is:
Substituting for f:
Newton’s second lawof motion:
The force of gravity is:
And simplifying
Initial conditions:
We get:
“ + c ”
Because it’sour initialcondition
Vector equation for ideal projectile motion:
Vector equation for ideal projectile motion:
Parametric equations for ideal projectile motion:
Rearranged
Horizontal component
Vertical component
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/sec
Or 741.1 miles/hr at sea level.
The projectile will be 2.5 kilometers downrange andat an altitude of 3.84 kilometers.
ParametricEquations
meters
meters
Horizontal component
Vertical component
Now, the maximum height of a projectile occurswhen the vertical velocity equals zero.
time at maximum height
This makes sensebecause the path ofa projectile is aparabola so themaximum wouldoccur at the vertex
Because object is not going up anymore atthis point
Recall:
Then,
Velocity
The maximum height of a projectile occurs whenthe vertical velocity equals zero.
We can substitute this expression into the formulafor height to get the maximum height.
This is theheight becauseit is the verticalcomponent
Simplifying, we get,
maximum
height
Andmultiplyingby “1”,
we get:
When the height is zero:
1) The time at launch:
Vertical component
Now, if we factor out t we have:
For this to equal 0, two things can happen:
OR
When the height is zero:
time at impact
(flight time)
Which gives us the
If we take the expression for flight time (time atimpact) and substitute it into the equation for x, we canfind the range.
If we take the expression for flight time and substituteit into the equation for x, we can find the range.
Range
(Simplifying)
The range is maximum when
Range
is maximum.
Range is maximumwhen the launchangle is 45o.
Recall:
If we start with the parametric equations for projectilemotion, we can eliminate t to get y as a function of x.
If we start with the parametric equations for projectilemotion, we can eliminate t to get y as a function of x.
This simplifies to:
which is the equation of a parabola since
it is a quadratic function.
If we start somewhere besides the origin, the equationsbecome:
As opposed to the parametricequations for ideal projectilemotion:
Example 4:
A baseball is hit from 3 feet above the ground with aninitial velocity of 152 ft/sec at an angle of 20o from thehorizontal. A gust of wind adds a component of -8.8 ft/secin the horizontal direction to the initial velocity.
The parametric equations become:
These equations can be graphed on the TI-89 to modelthe path of the ball:
Note that the calculator is in degrees.
t2
Max height about 45 ft
Distance traveled about 442 ft
Time
about
3.3 sec
Using
the
trace
function:
In real life, there are other forces on the object. Themost obvious is air resistance.
If the drag due to air resistance is proportional tothe velocity:
(Drag is in the oppositedirection as velocity.)
Equations for the motion of a projectile with linear dragforce are given on page 546.
You are not responsible for memorizing these formulas.