A competitive diver makes a dive from a platform 20 feet above the water. She reaches the top of her trajectory 2 feet above her starting point, and 2 feet out, horizontally from her starting point.
a. What’s the equation that describes her trajectory?
b. What’s the horizontal distance traveled when she hits the water?
+36915 Parabola with vertex at 2, 22
Find her initial vertical speed (she only goes up 2 feet against gravity)
vf = 0 = v0 + at a = - 32.2 ft/s^2
0 = vo - 32.1 t shows vo = 32.1 t
then d = do + vo t - 16.1 t^2 <===== Use this to find 't'
22 = 20 + (32.1 t) ^t - 16.1 t^2
2 = 16.1 t^2
t = .3525 seconds her horizontal speed is then 2 ft / .3525 s = 5.675 f/s and vertical speed is 32.1 * .3525 = 11.31 f/s
Find a point on the parabola
at t = 1 sec x = 5.675 ft and y = d = do + vo t + 1/2 at^2 = 20 + 11.31(1) - 16.1 (1^2) = 15.21 ft
vertex form of the parabola
y = a ( x-2)^2 + 22 and when x = 5.675 y = 15.21
shows a = -.503
a) Trajectory equation becomes y = -.503 (x-2)^2 + 22
b) when she hits the water y = 0 = -.503 ( x-2)^2 + 22 shows x = 8.6 ft
Here is a graph that shows this:
