SOLUTION: Part of a cheerleading routine is throwing a flyer straight up and catching her on the way down. The flyer begins this move with her center of gravity 4 feet above the ground. She

Question 1077367: Part of a cheerleading routine is throwing a flyer straight up and catching her on the way down. The flyer begins this move with her center of gravity 4 feet above the ground. She is thrown with an initial vertical velocity of 30 feet per second.
1.) Will the flyer�s center of gravity ever reach 20 feet? In two or more complete sentences, explain how you found your answer.
2.) For the flyer to have her center of gravity reach 25 feet, what does the initial velocity need to be? In two or more complete sentences, explain how you found your answer.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Part of a cheerleading routine is throwing a flyer straight up and catching her on the way down. The flyer begins this move with her center of gravity 4 feet above the ground. She is thrown with an initial vertical velocity of 30 feet per second.
You didn't give a function.
h(t) = -16t^2 + v0*t + h0 is commonly used for feet.
--> h(t) = -16t^2 + 30t + 4
=================
1.) Will the flyer�s center of gravity ever reach 20 feet? In two or more complete sentences, explain how you found your answer.
The max height is the vertex of the parabola at t = -b/2a
t = -30/-32 = 15/16 seconds
h(15/16) = -14.0625 + 28.125 + 4 = 18.0625 feet
Less than 20 feet
===================
2.) For the flyer to have her center of gravity reach 25 feet, what does the initial velocity need to be? In two or more complete sentences, explain how you found your answer.
The vertex is 25 feet.
The launch speed is the same as the speed descending.
Find the time to fall from 25 feet to 4 feet:
Falling from 25 feet:
h(t) = -16t^2 + 25 = 4
16t^2 = 21
t = sqrt(21)/4 seconds
---
v = at = 32*sqrt(21)/4 = 8*sqrt(21)
v = 36.66 ft/second

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