What's faster? Going up or Coming Down?

Suppose you are able to shoot a ball vertically straight up in the air and observe its motion (I don't know why... you might have a ballistic canon installed in the ground that allows you to do such a thing).

Assuming there are no crosswinds or other disturbances, you would expect its trajectory to be straight up, slow to a stop, and come straight back down.

But what takes longer? For the ball to reach its maximum height from when it was shot? Or for the ball to fall from its maximum height to the ground? Your intuition might suggest that it takes longer for the ball to reach max height, right? Anyway, it's a 50/50 guess I suppose.

Well, thanks to Sir Isaac Newton and the wonderful field of study called Calculus, we don't have to guess. We can work it out!

Figure 1.

To keep this relatively simple, let's assume that the only forces acting on the ball after it's been shot are retarding air resistance and the force of gravity. Suppose air resistance is linearly proportional to the ball's velocity and opposing it's motion. Thus...

And force do to gravity is simply the ball's weight (mass times acceleration due to gravity)...

The free-body diagram depicting this phenomenon is shown Figure 1 above. The blue arrow denotes the initial velocity, immediately after the ball is shot. The red arrows denote forces acting against the ball's motion.

Let's find a solution for the velocity and position of the ball as a function of time t, because we'll need both of these solutions to determine the answer to this (pointless?) question...

Now, if we sum up all of the forces on the ball in the vertical direction y, this is equal to the mass of the ball times its acceleration, according to Newton's Second Law of motion. Thus...

(Note: Σ means "Sum of")

Now we can express the acceleration of the ball as a derivative of its velocity with respect to time, i.e. a = dv/dt. Thus equation (1) becomes...

Now, equation (2) is a first-order differential equation, for which we can separate the variables as shown in equation (3)...

Integrating both sides and solving, we get...

Note: in equation (4) A = e^C, an arbitrary constant of integration for the general solution. We can find a particular solution by applying the initial conditions. When the ball is launched at t = 0...

Substituting the solution for A into (4), we get...

Now that we have an expression for velocity of the ball as a function of time, we can realise that it is the derivative of its position y, i.e. v = dy/dt. And thus we can turn this into another integral problem to find the position as a function of time.

We can find the arbitrary integration constant D by again, applying the initial condition. For completeness, let's take the initial position of ball at t = 0 as ...

Substituting the above express for D in to equation (6), we get...

That's a long expression, but equation (7) gives us the position of the ball at any time from when it is shot.

Now, at the top of its trajectory, when the ball reaches its maximum height, it will slow to a stop. Thus we simply need to set v = 0 to find the time when it reaches the top...

Ok, so we've got all of the equations we need in terms of symbols and algebra. Let's apply some real numbers so get can really see whether or not it is faster to go up than to come down or vice versa.

Say we have the following parameters...

Mass of the ball m = 1kg
Initial velocity = 20m/s
Initial position = 0
Air resistance coefficient p = 0.1
Acceleration due to gravity g = 9.81m/s²

Therefore, from equation (5), the particular solution for velocity is...

From equation (7)...

From equation (8)...

Trajectory from equation (10) is graphed in Figure 2 along with . The max height is y₁ = 17.98m.

Figure 2.

It appears that the trajectory is parabolic, which implies that it is symmetrical about the vertical dashed line .... But is it...?

Obviously the first solution to y(t) = 0 is t = 0. Unfortunately it is impossible to explicitly find the other solution from equation (10), so I've just used a "guess-and-check" approach (not very efficient - I might add) to find that y(t) crosses the horizontal axis at t = 3.83s, at which time the ball lands.

But if y(t) was symmetrical about t₁, the ball would land at 2t₁ = 3.71s, which is 0.12s sooner than y(t) indicates. This is shown in Figure 3 below...

Figure 3.

Figure 3 also includes the velocity curve of equation (9). The ball lands with a speed of 17.6m/s.

So, after all of that, we can conclude it is slightly quicker for a ball to go up than it is for it to come down. This variance was due to the inclusion of air resistance in our model. If this motion was conducted in a vacuum, then the trajectory would have been parabolic, and it would have taken an equal amount of time for the ball to go up and to come down.

A comparison between a with air resistance (yellow trajectory) and without air resistance (blue trajectory) is shown in Figure 4. As you can see, the ball reaches a significantly higher height (over 2m more) and is "airborne" for a bit longer without any air resistance.

And this result makes sense, because air resistance is always acting against the motion of the ball, whereas the force due to gravity is constant and always acting in the downward motion.

So on the way up, there are 2 forces acting downwards on the ball, limiting its overall vertical rise. Whilst on the way down, there is one force in the upward direction which increases as the ball gets faster (air resistance) and one constant force in the downward direction (gravity).

On a final note, we could have also approached this entire problem by realising that equation (2) is a linear first-order equation and used an integrating factor.

Credits:

This is my worked solution from an Applied Project from the book Calculus 5th Edition by James Stewart.

All equations in this tutorial were created with QuickLatex

All graphs were created with www.desmos.com/calculator