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Difficulty: EasyQuadratic Equations

A toy rocket is launched vertically upward from ground level. Its height, hh, in meters, tt seconds after launch is modeled by the equation h(t)=−5t2+40th(t) = -5t^2 + 40t. How many seconds after launch does the rocket reach its maximum height?

  1. A
    8
  2. 4Answer
  3. C
    80
  4. D
    20

Answer

4 seconds
The height of the rocket is modeled by the quadratic function h(t)=−5t2+40th(t) = -5t^2 + 40t. Because the coefficient of t2t^2 is negative (a=−5a = -5), the graph of this function is a parabola that opens downward, meaning its vertex represents the maximum height. The time tt at which the vertex occurs is given by t=−b2at = -\frac{b}{2a}. Substituting a=−5a = -5 and b=40b = 40 gives t=−402(−5)=4t = -\frac{40}{2(-5)} = 4 seconds.

Step-by-Step Solution

1
Identify the coefficients of the quadratic equation.
a=−5a = -5, b=40b = 40, and c=0c = 0
The equation is given in standard quadratic form h(t)=at2+bt+ch(t) = at^2 + bt + c.
2
Apply the vertex formula to find the time at which the maximum height occurs.
t=−b2at = -\frac{b}{2a}
For a downward-opening parabola, the maximum value occurs at the vertex.
3
Substitute the values of aa and bb and simplify.
t=−402(−5)=4t = -\frac{40}{2(-5)} = 4
This calculation yields the time in seconds when the rocket reaches its maximum height.

Key Concept

Finding the vertex of a quadratic function to determine the maximum or minimum value in context.

Alternative Method

Find the times when the rocket is at ground level by solving h(t)=0h(t) = 0. Factoring −5t2+40t=0-5t^2 + 40t = 0 gives −5t(t−8)=0-5t(t - 8) = 0, so the rocket is on the ground at t=0t = 0 seconds and t=8t = 8 seconds. Since a parabola is symmetric, the maximum height must occur exactly halfway between the launch and landing times, which is 0+82=4\frac{0 + 8}{2} = 4 seconds.
Estimated Time:45s
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Topics

Quadratic Equations

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