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Question

If a ball is thrown directly upward with a velocity of 22 ft/s, its height (in feet) after t seconds is given by y = 22t − 16t2. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.

Answer 1

Here you have a parabolic function that opens downward modeling the height of the ball. The maximum height will be achieved at the vertex, since after that the ball starts falling back down, so we need to find the vertex!

 

Rearranging the equation into a standard form gives us: y = -16t² + 22t + 0

 

Our equation to find the x-value of the vertex (or in this case t) is -b / 2a. So we just plug in our values from the equation and get:

t = -22 / (2 * -16) = -22 / -32 = 11 / 16

 

To find the y value (which will be the height of the ball at that time) we plug what we got for t back into the original equation:

y = -16(11 / 16)² + 22(11 / 16)

y = -121 / 16 + 242 / 16

y = 121 / 16 feet

 

This can be answered as is, or evaluated to a value of about 7.56 feet.

Answer 2

The equation given, if changed into standard form, is y = -16t² + 22t + 0.  It's an inverted parabola, opening downward, so the maximum height will occur at the vertex.

 

The vertex of a parabola is located at -b/2a, here -22/-32 = .6875.  Plugging this into our equation gives

y = -16(.6875²) + 22(.6875) = -7.5625 + 15.125 = 7.5625.

Rounding to the nearest whole number gives 8 feet.