Physics 101, 2004: Chapter 3

1
Chapter 3.3 (continued) 2D acceleration
vector acceleration
Instantaneous acceleration ?t?0.

• An object is accelerating if
• The magnitude of its velocity (i.e., speed)
  changes but its direction of motion stays the
  same.
• The direction of motion changes but its speed
  stays the
• same.
• Both magnitude of velocity and direction change.

2
Chapter 3.4 Projectile motion

• When an object moves in the xy plane,
• If the acceleration is uniform (usually downward)
  we call this projectile motion.
• We can think about its x and y motions
  independently. NOT OBVIOUS!

In our analysis we will always neglect air
resistance. Good approximation if velocities not
too large. For a projectile fired by a gun, its
a rather poor approximation.
3
Rules of projectile motion

• The x- and y-directions of motion can be treated
  independently.
• The x-direction is uniform motion
• ax 0
• The y-direction is free fall
• ay -g
• The initial velocity can be broken down into its
  x- and y-components.

4
General kinematical equations for constant
acceleration in two dimensions
Kinematical equations for projectile motion. ax
0 and ay g 9.80 m/s2.
x component (horizontal)
y component (vertical)
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Example A movie stunt driver on a motorcycle
speeds horizontally off a 50.0 m high cliff. How
fast must the motorcycle leave the clifftop if it
has to land on level ground below, 90.0 m
from the base of the cliff?
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More on Projectile Motion
What geometrical form does the trajectory
have? Assume that air resistance can be neglected.
Set x00, y00, and t00.
We want to eliminate t to see how x is related to
y. Easiest to solve the x-equation for t
Quadratic! Is parabola.
7
Calculating range of a projectile
Try to calculate the range R of a projectile
given initial conditions, v0 and T. Again set
x00, y00, t00.
Think of x and y motion separately y (vertical,
free fall) motion determines how long it takes
the projectile to drop back to the ground. Lets
get this time t.
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Projectile is launched at y00 and drops back to
y0.
Has two solutions t0 t2v0y/g
Solve for the time t.
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Range is equal to the distances traveled
horizontally at constant velocity v0x
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Now we can express the range, R, as a function of
the launch velocity, v, and the aiming angle, T
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The quite simple final result is
Maximum range is at angle of 45 degrees.
12
Example A fire hose held near the ground shoots
water at a speed of 6.5 m/s. At what angle(s)
should the nozzle point in order that the water
land 2.0 m away. Why are there two
different angles?
13
Example A 2.00 m tall basketball player wants
to make a basket from a distance of 10.0 m. If he
shoots the ball at a 450 angle, at what initial
speed must he throw the ball so that it goes
through the hoop without striking the backboard?