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 + 6
3t + 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?
2 answers:
Answer:
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.
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