Question

1) A distress flare is shot straight up from a ship’s bridge 75 ft above the water with an initial
velocity of 76 ft/sec.

A) Write a set of parametric equations for the path of the flare.
What window gives you the best picture of the problem situation?

B) What is the height of the flare after 2 seconds?

Answer 1

Parametric equations just require you to define a function of two variables (typically x & y) in terms of a third (ex t). Since this flare is assumed to be moving in two dimensions, parametrization of an x, y equation will suffice.

How can you think about the x position for any given t, since the x is always 0, for any change in t, its function is very simple. x(t) = ?

For its y position, you need to recall a general motion equation:
\[p(t) = p_0 + v_0(t) + \frac{ 1 }{ 2 }a(t)^2\] This is the general one. If you are solving for y position \[p_y(t) = p_{y0} + v_{y0}(t) + \frac{ 1 }{ 2 }a_{y}(t)^2\] this is the more specific version.

Answer 2

*you need to understand that a = -9.8 (a positive value of gravity, would mean the object is accelerating up into the sky)

Answer 3

what do i write for the parametric equations? I don't know how to do that @cdosborn

Answer 4

They will take the shape, x(t) = ... and y(t) = ... remember that before parametrization you would have y(x) = ... however can you figure out what the x(t) would be?

Answer 5

mmm idk \: I don't really have any idea what I'm doing, I just know I'm about to fail this class. @cdosborn

Answer 6

You're best chance is to seek understanding in these problems, not just the solutions. You're trying to find the path, which means position. Parametrization (introducing another parameter) says can we think of our original variables in terms of another variable. But simply put, were generally thinking about x position in terms of time and y position in terms of time. What is the x position at t=0, how about at t=3, or t=5?