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?
For the flyer to have her center of gravity reach 25 feet, what does the initial velocity need to be?

Question

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? 
    For the flyer to have her center of gravity reach 25 feet, what does the initial velocity need to be?

luffingsails · Answer

So, to determine how high the cheerleader goes, we need to solve the formula
$$Vf^2 - Vo^2 = 2gh$$
Where Vf = velocity of the cheerleader at the highest point
Vo = initial velocity
g = gravity constant
h = height traveled

At the highest point, the cheerleader will have stopped traveling, so Vf = 0.

$$0^2 - 30^2 = 2 *(- 32.15) * h$$
$$-900 = -64.3 * h$$
h = -900 / -64.3 = 13.99 feet.

Add the initial height of 4 feet and we have 17.99 feet. So, no the cheerleader will not make 20 feet.

luffingsails · Answer

To solve the second part, just set h = 25 - 4 = total height - initial height = 21. You know everything except for Vo, so just solve for that.