Question

A ball is thrown directly upward from height of 2 ft with an initial velocity of 24 ft/sec. The function is s(t)=-16 t^2+24t+2 give the height of the ball, in feet, t seconds after it has been thrown. Determine the time at which the ball reflects its maximum height and find the maximum height.

Answer 1

• time: 0.75 seconds
• height: 11 feet

Explanation:

A graphing calculator shows you the maximum height of 11 feet is reached after 0.75 seconds.

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A quadratic of the form ax² +bx +c has its vertex at x=-b/(2a). Here that means the time to the maximum height is ...

t = -24/(2(-16)) = 24/32 = 3/4 . . . seconds to maximum height

The height at that time is ...

s(3/4) = (-16(3/4) +24)(3/4) +2 = (-12 +24)(3/4) +2 = 9 +2

s(3/4) = 11 . . . . feet at maximum height