Calculus Gravity 3
Calculus Gravity 3
PROJECT REAL WORLD:
WHICH IS FASTER, GOING UP OR
By Moccia, Ariel, Stephanie, Taryn,
A ball with a mass (m) is projected upward from
Earth’s surface with a positive initial velocity (V0).
We assume the forces acting on the ball are the
force of gravity and a retarding force of air
resistances with direction opposite to the direction
of motion and with magnitude p|v(t)|, where p is a
positive constant and v(t) is the velocity of the ball
at time t. In both the ascent and the descent, the
total force acting on the ball is –pv-mg.
Confusing right? Well this is what we found from this complicated problem
•Air resistance and gravity are both acting upon the ball (not in a vacuum).
•G=Acceleration due to gravity
•P=Positive constant (air resistance)
•-pv-mg= The total force acting on the ball, ascending and descending
Part One: Prove that V(t) = (v0 + mg/p)ept/m -mg/p is the
differential equation of mv = -pv-mg
1. Separate unlike
terms, so that dv is
with v and dt is
2. Now we are going to
find the integral of
3. Take out the m.
4. Notice what U is
(the inner function),
and what du/dv is
(the derivative of
5. Multiply in a ‘-p’ so
you can use the
method, do not to
put divide by a -p
outside of the
6. Use the substitution
method now, and
Find the antiderivative of
the integrals now.
Isolate the ln(-pv-mg) by
multiplying the other side
by the reciprocal of –m/p.
(pt/m)(C) is just another
There is a rule for ln that
says if ln(a)=b, then a= eb.
So, -pv-mg= e-pt/m + D.
We can separate e-pt/m + D
into e-pt/m eD. eD is just
another constant, so we
can call it K. Thus, we get
Now we have to find what
K, the constant, is equal
to. The time (t) is 0, so we
can plug that in, e0 = 1, so
we have isolated K.
Plug that back into the
equation where K is.
Isolate V, and after you
can see that V(t)= (V0+
Part Two: Show that the height of the ball, until it hits
the ground is y(t)= (V0 + mg/p) m/p (1 – e-pt/m) –mgt/p
1. The derivative of the velocity equation will
give us the height.
2. Find the integral of dy and the integral of
the other side of the equation.
3. The derivative of e-pt/m is e-pt/m/-p/m. This is
the same thing as e-pt/m × (-m/p).
4. Mg/p now become mgt/p, because the
antiderivative of t0 is t1/1.
5. Even though you have Y now, you still have
a constant, so solve for the constant (C).
6. Plug C back into the equation.
7. Take out like terms.
8. Now you have the equation for height.
Part Three: If t1 is the time that the ball takes to reach its maximum
height, show that t1= m/p ln[(mg+pv0)/mg]. Then find the time for a
ball with a mass of 1 kg and an initial velocity of 20 m/s, and with
air resistance being 1/10 of the speed.
Set the equation of v(t)
equal to zero, because
at the maximum height
the ball stops for a
second (thus the
velocity is zero). This is
so we can solve for t1.
Isolate the part that has
t in it, so make sure that
e-pt/m is isolated.
Using lna = B, a=eb, we
see –pt/m is equal to ln(
mg/pv0 + mg).
Use the rule rloga =
Logar to solve for t.
V1 = 20 m/s, p = 1/10, m= 1 kg.
Plug in all the given values to
Part Four: Let t2 be the time at which the ball falls back to Earth. For
the same ball in Problem 3, estimate t2 by using the graph of the
height function (y). Which is faster, going up or coming down?
If it goes the same speed
going up as it does coming
down, then the time will
be the same. Thus 2t will
be the total time.
Plug in 2t (3.71) into the
y(t) equation, and solve for
y. The y will tell you how
far from reaching the
ground the ball is.
The ball is faster going up
than it is coming down!