Video Transcript
An artillery shell is fired
vertically upward at a target 200 meters vertically above the ground. When the shell is 100 meters above
the ground, it has a speed of 100 meters per second. What is the speed of the shell when
it hits the target? Neglect drag forces on the
shell.
We can name the speed that the
shell has when it hits the target 𝑣 sub 𝑇 and begin in our solution by drawing a
diagram of this process. In this example, we have a
vertically pointed artillery canon firing a shell straight up. When the shell passes the 100-meter
elevation mark, it’s moving at 100 meters per second. We want to know, when it reaches
its target at 200 meters, what is its velocity, 𝑣 sub 𝑇. In this scenario, we can define
vertical motion as motion beginning at ground level and positive in the upward
direction. When the shell leaves the canon, it
will be subject to only one force, the force of gravity acting down on it. Because of our sign convention, we
write 𝑔 as negative 9.8 meters per second squared. Since the artillery shell is
accelerating constantly, due only to the effects of 𝑔, we know that the kinematic
equations apply to this scenario.
Looking over these equations, in
terms of the variables they’re written in, we want to solve for a final velocity 𝑣
sub 𝑓. That’s what 𝑣 sub 𝑇
represents. Both of the first two equations
written show that variable. So let’s see if there’s more
information that can help us choose between these two. Looking at our diagram, based on
our problem statement, we see that we know distances in the vertical direction. And we also know the acceleration
acting on the shell. When it comes to the variables
involved, the only difference between the first and the second kinematic equation
have to do with time versus distance. Since we’re given distance, but not
time, let’s choose to work with the second equation.
Writing it in terms of our
variables, we can say that 𝑣 sub 𝑇 squared is equal to 𝑣, the velocity of the
shell when it’s at 100 meters elevation, plus two times 𝑔 times 𝑑. Where 𝑑 is the distance between
that 100- and 200-meter mark. Taking the square root of both
sides of the equation and given 𝑣 and 𝑔 and recognizing that 𝑑 must be equal to
100 meters, we’re ready to plug in and solve for 𝑣 sub 𝑇. With these values plugged in,
notice how important the signs are, that we established a sign convention and
followed that with these values. It’s the sign convention that tells
us that 𝑣 is positive, 𝑔 is negative, and 𝑑 is positive. When we calculate this expression,
we find that, to three significant figures, it’s 89.7 meters per second. That’s the velocity of the shell in
the upward direction when it reaches the target.
