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The height of a ball thrown in the air can be described by a parabola. After 3

seconds, the ball reaches the maximum height of 42 feet. If the ball was originally
thrown from a height of 6 feet, find an equation for the height, h(t), of the ball after t
seconds.

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Answer:In his equation height is given by h(t) and t represents time. The shape of this equation is that of a downward facing parabola because the coefficient of the first term is negative. Therefore, the maximum height is found by determining the vertex of the parabola which is given by b/2a or 50/(16*2). Thus after, 1.5625 seconds the ball reaches it maximum height of 44 feet (plug 1.5625 into formula.)

Notice that the formula has a constant of 5 in it meaning that the ball was throw up at a 5 foot starting position above the ground. Thus, for the ball to hit the ground h(t) = -5. So, what is the value of t that gives that value? Here we just use the quadratic formula and find that t = 3.31 seconds. That makes sense doesn't it? 2*1.5625 would get the ball back to the starting position ( 3.125 sec) and the ball needs another .185 sec to fall the last 5 feet. What speed is that in MPH? That would be 5/.185*(3600/5280) = 18.4 MPH. That is a lot slower than a 95 MPH fast ball.

Explanation:

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