A ball is thrown into the air from a height of 4 feet at time t = 0. The function that models this situation is h (t) = - 16t2 + 63t + 4, where t is measured in seconds and h is the height in feet. a) What is the height of the ball after 3 seconds? b) What is the maximum height of the ball? Round to the nearest foot. c) When will the ball hit the ground? d) What domain makes sense for the function?

a) 49 ft

b) 66 ft

c) 4 seconds

d) [0, 4] seconds

Step-by-step explanation:

a) Evaluate the function for t=3:

h (3) = - 16·3² + 63·3 + 4 = (-16·3 + 63) ·3 + 4 = 15·3 + 4

h (3) = 49

The height of the ball is 49 feet after 3 seconds.

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b) The maximum height of the ball will be found where t=-b / (2a) = - 63/-32 = 1.96875.

h (1.96875) = (-16· (63/32) + 63) · (63/32) + 4 = 63²/64 + 4 = 66.015625

The maximum height of the ball is approximately 66 feet.

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c) The ball will hit the ground when its height is zero.

-16t² + 63t + 4 = 0

Using the quadratic formula, we find the solution to be ...

t = (-63 - √ (63² - 4 (-16) (4))) / (2· (-16)) = (-63 - √4225) / -32 = - 128/-32 = 4

The ball will hit the ground after 4 seconds.

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d) The function is only useful for the time period between when the ball is thrown and when it lands, t = 0 to t = 4 seconds.

The domain of t in the interval 0 to 4 seconds makes sense for this function.