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 886833: 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.
So knowing this:
Will the flyer�s center of gravity ever reach 20 feet?
For the flyer to have her center of gravity reach 25 feet, what does the initial velocity need to be? Thanks.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
h(t) = -16t^2 +(vo)(t) + ho, where vo=velocity in ft/sec and ho= initial height in ft
we are given vo = 30 and ho = 4, therefore
h(t) = -16t^2 +30t +4
now we know that the x coordinate of the max altitude is -b/2a and in our problem
t = -b/2a = -30/(2*-16) = 0.9375
so max height is reached in 0.9375 seconds, now we can calculate the max height
h(0.9375) = -16*(0.9375^2) +30*0.9375 +4
h(0.9375) = 18.0625 feet
The flyer's center of gravity does not reach 20 feet
now find initial velocity to reach 25 feet
25 = -16*(0.9375^2) +vo*0.9375 +4
25 = −14.0625 +vo*0.9375 +4
vo*0.9375 = 35.0625
vo = 37.4 feet per second
therefore the initial velocity must be 37.4 feet per second to reach a height of 25 feet

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