Algebra 2 Throw Formula Problem

[MasonD]

In my Algebra 2 class, we are working on a project in which we built a catapult and took measurements of it's launch to model a quadratic. In one of the problems, I have to use the throw formula (h=-16t^2 + vt + s (v = velocity[ft/sec], s = initial height[feet], h = height[feet], t = time[seconds])) to create a quadratic model. The table of values I recorded for the catapult launch is below:

	Distance(cm)	Maximum Height(cm)	Time(sec)
Trial 1	138	114	1.32
Trial 2	142	103	1.06
Trial 3	152	107	1.15
Average	144	108	1.176

My teacher wants me to use average values. I know that I have to convert the centimeters into feet and then divide by the time, but what would I do after that?

 

[Otis]

… we built a catapult and took measurements of [its launches] to model [the projectile motion with] a quadratic [equation] …
h = -16t^2 + vt + s …

Hello Mason. In that formula, symbol v represents the vertical component of velocity (at the instant the object is released by the catapult). The object's initial velocity has also a horizontal component. Have you been modeling both the vertical and horizontal motion, in class?

I'm thinking your first column (labeled 'Distance') gives the horizontal distance traveled. Is that correct? If it is, then can you state the formula for horizontal distance (d) in terms of time?

Also, what is the average initial height for those launches?

… I know that I have to convert the centimeters into feet and then divide by the time …

When you say that you need to divide by time, are you thinking about the vertical velocity's units (ft/sec)? That velocity is number of feet per one second. You don't need to divide the number of feet by 1. Otherwise, please explain the division that you're considering.

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[MasonD]

For the project, we also had to make a model of height as y and distance as x. I'm not sure what you mean when you say "the formula for horizontal distance in terms of time." My teacher said we are just using 0cm as the initial height for the launches. When I said dividing by time, I meant dividing the average height of 108cm (converted to feet) by the average time (divided by two because the time is when the projectile hits the floor. I would divide by two to see at what time the projectile is at a height of 108cm [the vertex]). It would look something like: 3.54(108cm in feet) / 0.588(1.176 seconds divided by two). I hope that makes sense.

 

[Otis]

… we also had to make a model of height as y and distance as x … we are just using 0cm as the initial height …

Okay, so it seems for now that you're interested only in finding the value of v, to complete the quadratic model for height in terms of time. That is, we don't need the distance column. The averaged maximum height and flight time are sufficient.

You're correct about the maximum height being reached halfway through the flight (approximately, in this project). That corresponds to the vertex of the parabola lying on the parabola's axis of symmetry (which intersects the horizontal coordinate axis halfway between the intercepts).

Therefore, we can use that point in time (when the object is at maximum height, approximately) to find the value of v because we know the values of h, t and s then.

h = -16t^2 + v*t + s

First, substitute the given value for s.

Next, substitute the average maximum height for h (3.54).

Now substitute half of the total time (0.588) for t.

Solve the resulting equation for v.

PS: By ignoring the true initial height, I suspect it will be a somewhat sloppy model.

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