Download presentation
Presentation is loading. Please wait.
1
10.4 Projectile Motion Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2002 Fort Pulaski, GA
2
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)
3
We assume that the projectile is launched from the origin at time t =0 with initial velocity v o. The initial position is: Horizontal component Vertical component At the origin Note: We will see this again
4
Newton’s second law of motion: Vertical acceleration Force = Mass (Acceleration) Second derivative = acceleration
5
Newton’s second law of motion: The force of gravity is: Force is negative Because gravity pulls downward
6
Newton’s second law of motion: The force of gravity is: Substituting for f:
7
Newton’s second law of motion: The force of gravity is: And simplifying
8
Initial conditions: We get: “ + c ” Because it’s our initial condition
9
Vector equation for ideal projectile motion:
10
Parametric equations for ideal projectile motion: Rearranged Horizontal component Vertical component
11
Example 1: A projectile is fired at 60 o 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 and at an altitude of 3.84 kilometers. Parametric Equations meters Horizontal component Vertical component
12
Now, the maximum height of a projectile occurs when the vertical velocity equals zero. time at maximum height This makes sense because the path of a projectile is a parabola so the maximum would occur at the vertex Because object is not going up anymore at this point Recall: Then, Velocity
13
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.
14
This is the height because it is the vertical component Simplifying, we get,
15
maximum height And multiplying by “1”, we get:
16
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
17
When the height is zero: time at impact (flight time) Which gives us the
18
If we take the expression for flight time (time at impact) and substitute it into the equation for x, we can find the range.
19
If we take the expression for flight time and substitute it into the equation for x, we can find the range. Range (Simplifying)
20
The range is maximum when Range is maximum. Range is maximum when the launch angle is 45 o. Recall:
21
If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.
22
This simplifies to: which is the equation of a parabola since it is a quadratic function.
23
If we start somewhere besides the origin, the equations become: As opposed to the parametric equations for ideal projectile motion:
24
Example 4: A baseball is hit from 3 feet above the ground with an initial velocity of 152 ft/sec at an angle of 20 o 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:
25
These equations can be graphed on the TI-89 to model the path of the ball: Note that the calculator is in degrees. t2t2
27
Max height about 45 ft Distance traveled about 442 ft Time about 3.3 sec Using the trace function:
28
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: (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.
Similar presentations