3-4 Special Case 1: Projectile Motion
When you launch a projectile, you have to give it an initial velocity to get it
going. Because this is projectile motion, let’s assume the object is launched at
some angle, θ. Once you impart that initial velocity, the object takes off. You
do NOT keep applying that initial velocity.
With regard to that velocity,
velocity two things can / do happen:
1. The initial horizontal velocity, v0x, is given as v0x = v0 cos θ. Because you
do not keep applying force to the object after that point, v0 stays
constant. Thus, there is no horizontal acceleration, to speak of. Actually,
horizontal velocity stays constant within a relatively short time frame. In
an extended time frame, with a very fast velocity, air resistance could
become a factor.
2. The vertical velocity does change with time. Initially, v0y is given to be
v0y = v0 sin θ. Right out of the box, the vertical acceleration is defined as
– g. Some resources, texts, will define g as positive; some as negative.
You must check your frame of reference. It’s negative here, meaning y
(or the height) will decrease with time. This means that the initial
vertical velocity would depend on the velocity with which it was
launched, as well as the effect on that velocity due to gravity. So, the
overall y-component of velocity will be the combination of the two, or,
vy = v0y + - gt. Unit-wise, this makes sense, as – gt will give you units for
velocity, overall, while implicitly stating that the y-component of velocity
of an object in free-fall does vary with time.
Take a look at the following snapshots of an object in free-fall motion
OH
With regard to displacement,
displacement two things can / do happen:
Where something is, or its displacement, depends on two factors; did it have an
initial horizontal velocity, and did it have an initial vertical velocity? If so, its
location is a function of both where it was initially, and where it can get to,
given time to do so.
1. For its displacement in terms of x, x(t) = x0 + v0xt
2. For its displacement in terms of y, y(t) = y0 + v0yt – ½ gt2
Just like before, unit-wise, this will give you units that agree with displacement.
Q: Why does the x-component of displacement have only two terms, while the
y-component has three?
Q: Why, specifically, do we add – ½ gt2 to the y-component of displacement?
SEE EX 3-8 P 66
NOTE: The normal y-versus-t scale has been converted to a y-versus-x scale
by multiplying the time values by 19.6, as the cap moves at 24.5 m/s cos 36.90 =
19.6 m/s, horizontally.
NOTE: For each angle and its complement, they have the same range.
Q: Which angle yields the maximum range?
Other Equations
1. Path of a Projectile - y wrt x:
y(x) = (tan θ0)x – x2(g/(2v02cos2θ0) See diagram 1 above
2. Range of a Projectile for Equal Initial and Final Elevations:
R = (v02/g)sin 2θ
θ0 See diagram 2 above
NOTE: Equation 1 comes from assuming x0 and y0 in the x(t) and y(t)
equations equal zero, solving for t, and substituting back into the y(t) equation.
Equation 2 comes from assuming y(t) = 0, solving for total time, and
substituting back in.
See diagram 3 above. ☺
SEE EX 3-9 P 68-69
Q: What do the positive and negative times mean?
YOU TRY IT. DO EX’S 3-10 – 3-12 ☺ White-board Activity!!
Pay especially close attention to EX 3-12. It is particularly germane wrt “Mean
Kitty.” You are now garnering even morr significant information regarding
the physics behind your “Mean Kitty” or “Shoot the Monkey” lab. You ought
to be in the stage where you are jotting down ideas, doing internet research,
and beginning to solidify a plan. Your apparatus must fit through a standard
door. Your labs will be due immediately after spring break, barring any other
unforeseen circumstances.