Physics Lab #6
Projectile Motion II
Theory
For an object traveling in earth's gravity in such a way that we can ignore air resistance, its horizontal (x) motion and vertical (y) motion is described by:
For position:
x = vocosθ t,
y = vosinθ t – ½g t2,
For velocity
vx = vocosθ,
vy = vosinθ – g t
For acceleration:
ax = 0 m/s2,
ay = – g = – 9.8 m/s2
where the initial position and time are chosen to be zero and the initial velocity is of magnitude vo (also known as launch speed) directed at angle θ above the horizontal. The horizontal direction is x and the vertical direction is y (up positive, down negative).
Using these equations we can show that the maximum height, H, the range (distance traveled horizontally if it begins and ends at the same height, i.e. the ground is flat), R, and the total time in the air, T are given by:
H= ½v2osin2θ/g,
R= v2osin2θ/g,
T= 2vosinθ/g
Remember from the
last funnelator lab:
From trigonometry we can figure out how tall something is, call it H, if we
know how far we are from the base of the object, call it d, and we measure the
angle, A, from the base to the top of the object. We get:
H = d tan(A)
Procedure
We will use the funnelator to shoot a rubber ball
upward in a parabolic trajectory. (Two bonus
marks for those who can use the first two equations in the theory section to prove
the shape is a parabola). We will use trigonometry to figure out the height of
the ball at maximum and then time the duration of the entire trip, from start
until it hits the ground. We will do a number of launches. Remember reading
errors.
Calculations
We will assume that our launch angles are accurate and
then calculate the launch speed from the H, R and T measurements. Compare results. Do they agree? Which is the most precise? From the first projectile lab, what misgivings
do you have about all this? How might
you fix these results? Also try to check
whether or not the maximum range was found for the expected angle.