math pre calc 11 please help :)?

[匿名]

2020-11-23 1:46 pm

really stuck on this question. was wondering if anyone could help asap!!

The path a basketball shot takes can be modelled by the function:
h(d)= -0.08d^2 +0.6d+2.1, where h(d) is the height (meters) the ball has travelled since it left the shooters hand.

a) What is the maximum height the basketball reaches?
b) How far has the ball travelled horizontally from where it was shot to when it reaches its maximum height?
c) What was the height of the ball when it was released?
d) The basket is approximately 3.08 meters off the ground. If the player made the shot, how far in meters from the
basket were they?

thank you so much:) really appreciated !!

回答 (1)

Uncle Michael

2020-11-23 3:40 pm

✔ 最佳答案

Since it is a pre-calculus course, no calculus would be used.

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a)
h(d) = -0.08d² + 0.6d + 2.1
h(d) = -0.08[d² - 7.5d] + 2.1
h(d) = -0.08[d² - 7.5d + 3.75²] + 2.1 + 0.08 × 3.75²
h(d) = 3.225 - 0.08(d - 3.75)²

For all real values of d, - 0.08(d - 3.75)² ≤ 0
Hence, h(d) = 3.225 - 0.08(d - 3.75)² ≤ 3.225
Maximum h(d) = 3.225
Maximum height = 3.225 m

====
b)
h(d) = 3.225 - 0.08(d - 3.75)² ≤ 3.225
Maximum h(d) when 0.08(d - 3.75)² = 0

When the ball reaches the maximum height,
distance travelled horizontally = 3.75 m

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c)
When the ball is released, d = 0
h(0) = -0.08(0)² + 0.6(0) + 2.1 = 2.1
When ball is released, height = 2.1 m

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d)
When h(d) = 3.08:
3.08 = -0.08d² + 0.6d + 2.1
0.08d² - 0.6d + 0.98 = 0
d = [0.6 ± √(0.6² - 4*0.08*0.98)]/2
d = 5.1 or d = 2.4
Distance = 5.1 m or 2.4 m

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