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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?

 Mar 26, 2023
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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:

 Mar 30, 2023

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