January 21, 2009

Let's solve our physics problem

It's time to solve that projectile motion physics problem I gave you last week.

Here's the problem (with my selection of 38.5 m as the distance from the kick point to the cross bar):

A ball is kicked from a point 38.5 m away from the crossbar. The top of the crossbar is 3.05 m high. If the ball leaves the ground with a speed of 20.4 m/s at an angle of 52.2º to the horizontal? (The usual assumptions apply: uniform earth gravity, no drag or wind, the ball is a point)

a. By how much does the ball clear or fall short of clearing the crossbar?

b. What is the vertical velocity of the ball at the time it reaches the crossbar?

Here's a graphic representation of the problem:

We know that the ball is kicked at an initial speed of 20.4 m/s at an angle of 52.2º. As such, the ball will be travelling both upwardly (vertical component) and to the right (horizontal component). Gravity will also be acting on the ball and pulling (accelerating) it back down to earth.

So, our problem is governed by the "what goes up, must come down" law of motion. In our case the ball gets kicked. It goes up. Then it comes back down. In between the going up and the going down parts of the trip, the ball reaches its maximum height.

We can easily calculate the maximum height (vertical distance travelled), the time it takes the ball to reach this maximum height, and the horizontal distance the ball travels during this time. Once we know these values we can determine where the crossbar is in relation to the maximum height reached by the ball. Does the ball pass the crossbar while it's going up or while it's going down? Both possibilities are shown in the picture.

Once we figure this out, then we'll worry about solving the rest of the problem.

So let's get to it and use some 21st Century skills. But, notice the amount of domain specific critical thinking that had to be done beforehand.

Let's google "projectile motion calculator" and select the first link.

Pay Dirt -- a projectile motion calculator presented as an HTML form. Filling out an HTML form seems to be a good example of a 21st Century skill, so let's use it.

The first two input fields call for an initial velocity (in m/s) and an angle (degrees). We know those values -- 20.4 and 52.2. Let's enter in those values and select Horizontal Component of initial velocity, Vertical Component of initial velocity, Max Height, Time to reach max height, Time of flight, and Range of flight as the things we want the calculator to calculate. The more the merrier. Computation time in the 21st century is cheap and this case nearly instantaneous.

Here's what the calculator spits out:

Horizontal Component of initial velocity: 12.503 m/s
Vertical Component of initial velocity: 16.119 m/s
Max Height: 13.256 m
Time to reach max height: t = 1.645 s
Total Time of flight = t = 3.290 s
Total distance travelled = 41.131 m

Wow. That saved me quite a bit of effort remembering and applying algebra, trigonometry, and physics. I think I like these fancy 21st Century skills.

Now let's put our critical thinking caps back on and make some sense of these numbers.

The calculator told us that the total time of flight (3.290 s) was merely double the time it took to reach the maximum height (1.645 s), i.e., it takes the same amount of time for the ball to travel up that it does to travel back down. This means that the distance to reach the maximum height is half of the total distance travelled 41.131 m ÷ 2 = 20.5655 m.

Since the cross bar is 38.5 m away from the starting point and the maximum height is reached at 20.5655 m, then the ball reaches its maximum height before it reaches the crossbar. So, the ball will be on its way back down to earth when it crosses the crossbar. This condition is shown in the picture and is the crossbar on the left hand side.

At the maximum height the ball is 13.256 m high and the crossbar is only 3.05 m high. The question is will the ball still be above the crossbar when it reaches the horizontal distance of the crossbar.

First we have to figure out how much farther the ball has to travel horizontally between the maximum height (20.5655 m) and the distance to the crossbar (38.5 m). This is simple subtraction (38.5 - 20.5655): 17.9345 m. The ball has 17.9345 m left to travel horizontally befoe it reaches the crossbar.

The calculator tells us that ball is travelling at a horizontal speed of 12.503 m/s.

I'm hoping that most people know that distance travelled = speed x time travelled. Or time travelled = distance travelled ÷ speed. In our case, the time it takes to traverse those 17.9345 m is (17.9345 m ÷ 12.503 m/s) 1.434 s.

Now the question is how far will the ball fall down during those 1.434 s.

That seems like real physics to me. So let's see if we can find another calculator to do the dirty work. As luck would have it the second link of our google search has just what we need. This page has a bunch of different projectile motion calculators. Again, as luck would have it, the top one "freefall" is the one we need (though you probably didn't know that unless you knew some physics).

The freefall calculator asks only for a time of travel. We just calculated that time as 1.434 s. Punching that number in, we find out that:

The vertical speed of the ball is 14.0532 m/s downward and the ball travels 10.076 m. So, the ball falls 10.076m from its max height of 13.25 m. so the ball is (13.25 m - 10.076 m) 3.174 m above the ground when it reaches the crossbar. The crossbar is only 3.05 m high, so the ball just barely clears the bar.

Now we have all the information we need to answer the problem.

Answer A: The ball clears the bar by (3.17 m - 3.05 m) 0.12 m.

Answer B: The ball's vertical velocity is 14.1 m/s downward according to the calculator.

We can check our answer by filling in our numbers in the "will it clear the fence" calculator toward the bottom of the page The answer does indeed check out.

Done and done.

We'll discuss the results in another post.


Anonymous said...

Yup... I am a Google Zen master and I gave up after 10 minutes. :)

Tracy W said...

I came up with a similar method and answer (well using my memory of trig and my beloved calculator, not an online one), but:
- I did physics at high school, back in the 20th century, and roughly remembered the procedure.
- It took me about 40 minutes, what with having to look up the speed laws and I haven't checked it yet.

Anonymous said...

I addressed this same issue here.