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. Will the flyer’s center of gravity ever reach 20 feet? In two or more complete sentences, explain how you found your answer. 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.

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AlonsoDehner AlonsoDehner · Answer 1 · 2018-12-04T09:10:32+01:00

Answer:

Step-by-step explanation:

Given that a flyer begins from earth and thrown up with a velocity of 30 ft/sec vertically.

u = initial velocity = 30 : a = -g = 32 ft/sec^2

s(0) =initial height =4 ft.

We have the equation

[tex]v^2 =u^2+2as[/tex]

where u = initial velocity : v= final velocity : s = distance travelled and a = acceleration. Here final velocity is found out as follows

Substitute to get

[tex]30^2-2(32)(20)\\ = 900-1280 =-380[/tex]

Since v cannot be negative, the flyer’s center of gravity cannot ever reach 20 feet.

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To reach 25 ft, put s = 25

and v must be atleast 0

Then we have [tex]0=u^2-2(32)(25)\\Or u^2=1600\\Or u =40[/tex]

Hence if thrown with initial velocity of 40 ft /sec, it will reach a height of 25 ft.