Position:
yinital = 100 m
xinital = 0 m (we are choosing the starting value to
bezero)
Velocity:
vx initial = 50 m/s
vy initial = 0 m/s
Acceleration:
ax = 0 m/s2 (we are neglecting
airresistance)
ay = g = -9.8m/s2
A Free Body Diagram couldn't hurt either:
I drew the velocity and acceleration components separate from
thecoyote so it didn't get cluttered. The arc represents thepath of
the coyote.
There are two parts to this question. First, we have todetermine
how long it takes for the coyote to hit the bottom. Then, we can
use that time to determine the distance the coyote hastraveled in
the horizontal component.
One of the key points to understanding this problem isunderstanding
that the horizontal and vertical components areindependent. That
is, the acceleration due to gravity is onlygoing to change the
velocity of the coyote in the verticaldirection. It will not affect
the velocity of the coyote inthe horizontal direction.
So, the first part: Determine the amount of time that ittakes the
coyote to hit the ground.
We can use the kinematics equations for this:
x = vavgt where vavg =
averagevelocity = (vfinal-vinitial)/2
x = v0t + 1/2 at2
v = v0 + at
v2 = v02+2ax
X is the displacement (change in position), v is the final
velocity(except in the first formula), v0 is the
initialvelocity, a is acceleration, and t is time.
So, which one do we use (most easily)? Think about what weare
looking for: time. That tells us to look at the topthree formulas
because the fourth doesn't have a "t" in it. Now look at all of the
other variables in the formulas and considerwhat we know and what
we don't know. We know the distancethat the coyote will fall, the
acceleration, and the initialvelocity in the vertical direction. We
don't know the finalvelocity that the coyote will have (just before
hitting theground). The second equation is the only one that
isindependent of final velocity, so we'll use that.
x = v0t + 1/2 at2
Changing the variables so as not to confuse it with the
horizontalcomponent:
y = v0yt + 1/2 ayt2
Now we can substitute in what we know:
-100 m = (0 m/s)(t) + 1/2
(-9.8m/s2)(t2)
And solve for t:
-100 m = 1/2 (-9.8m/s2)(t2)
-200 m = (-9.8m/s2)(t2)
-200 m / (-9.8m/s2) = t2
√(-200 m / (-9.8m/s2) ) = t
√(-200 m / (-9.8m/s2) ) = t
4.52s = t
Now we know how long the coyote will be traveling in the
horizontaldirection.
Using this time as well as the velocity in the horizontaldirection,
we can determine how far (horizontally) the coyote willland from
his starting point:
x = vt
x = (50 m/s)(4.52 s) = 226 m
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