中四數學題 20點 THX 幫explain

2014-10-27 12:35 am

題目: Ken throws a ball from the ground. It is given that the path of the ball is the graph of the function y = -x ^ (2) + 6x, where the vertical distance betwrrn the ball and the ground is y m, and the horizontal distance of the ball from Ken is x m.

(a)If Ken is standing 2.5m away from the building of 8m tall,can he throw the ball to the top of the building from the ground? Explain your answer.
答案係YES

(b)If Ken walks closer to the building until they are 1.5 apart, can he throw the ball to the top of the building from the ground? Explain your answer.
答案係NO

幫explain THX

回答 (1)

50418129

2014-10-27 5:24 pm

✔ 最佳答案

As the locus of the path of the ball is a parabola(opening downward), the vertical distance of ball will increase to the maximum point and then decrease.

Since y
= -x^2 + 6x
= -(x^2 - 6x)
= -[(x - 3)^2 - 9]
= -(x - 3)^2 + 9

So, when the horizontal distance of the ball is 3m, it reaches the maximum point which is 9m from the ground.

(a)

When Ken is standing 2.5m(<3m) from the building, the maximum vertical distance from the ground = -(2.5)^2 + 6(2.5) = 8.75m.
So, he can throw it to the top of the building.

(b)

When Ken is standing 1.5m(<3m) from the building, the maximum vertical distance from the ground = -(1.5)^2 + 6(1.5) = 6.75m.
So, he cannot throw it to the top of the building.

參考: knowledge

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