从一台机器上发射了一个物体。在发射后 tt 秒，它在地面上方的高度 yy（米）可以用二次函数 y=f(t)y = f(t) 来建模。发射后 4 秒，该物体达到了最大高度 81 米，并在再过 3 秒后下降到 54 米的高度。

  1. (a)

    求该物体被发射时的初始高度。

  2. (b)

    在不求解二次方程的情况下，解释并确定该物体在 54 米以上停留了多长时间？

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完整解法

1(a)

1. Formulate the quadratic function using the given maximum point

Since the maximum height is 81 m81\text{ m} at t=4t = 4, the vertex of the parabola is (4,81)(4, 81).
Let the quadratic function be y=a(t−4)2+81y = a(t - 4)^2 + 81.

2. Use the second given point to find the constant $a$

The object drops to 54 m54\text{ m} after a further 33 seconds, so at t=4+3=7t = 4 + 3 = 7, y=54y = 54.
Substitute (7,54)(7, 54) into the equation:
54=a(7−4)2+8154 = a(7 - 4)^2 + 81
54=9a+8154 = 9a + 81
9a=−279a = -27
a=−3a = -3
So, the function is y=−3(t−4)2+81y = -3(t - 4)^2 + 81.

3. Calculate the initial height by substituting $t = 0$

The initial height is when t=0t = 0:
y=−3(0−4)2+81y = -3(0 - 4)^2 + 81
y=−3(16)+81y = -3(16) + 81
y=−48+81=33y = -48 + 81 = 33

答案

33 m33\text{ m}

1(b)

1. Explain the use of symmetry to find the other time the object is at 54 m

The graph of a quadratic function is a parabola, which is symmetrical about its maximum point at t=4t = 4.
Since the object is at a height of 54 m at t=4+3=7t = 4 + 3 = 7 seconds, by symmetry, it must also be at 54 m at t=4−3=1t = 4 - 3 = 1 second.

2. Calculate the duration the object stayed above 54 m

Time spent above 54 m =7−1=6= 7 - 1 = 6 seconds.

答案

6 seconds

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