Question

NASA launches a rocket at t = 0 seconds. Suppose its height, in meters above sea-level, as a function of time is given by h = −4.9t^2 + 175t + 143. How high above sea-level does the rocket get at its peak? Round your answer to 2 decimal places.

The rocket peaks at ________ meters above sea-level.

Answer 1

Answer:

1,705.5 meters

Step-by-step explanation:

Given quadratic function:

$h(t) =- 4.9t^2 + 175t + 143$

As function h(t) is a quadratic function with a negative leading coefficient, the graph of h(t) forms a downward-opening parabola. Consequently, the maximum height reached by the rocket corresponds to the y-value of the vertex of this parabola.

The formula for the x-coordinate of the vertex of a parabola in the form y = ax² + bx + c is:

$x = -\dfrac{b}{2a}$

In this case:

• x = t
• a = -4.9
• b = 175
• c = 143

Substitute the values into the formula:

$t=-\dfrac{175}{2(-4.9)}\\\\\\t=-\dfrac{175}{-19.8}\\\\\\t=\dfrac{125}{7}$

Now, to find the maximum height, we can substitute this value of t into function h(t):

$h\left(\dfrac{125}{7}\right) = -4.9\left(\dfrac{125}{7}\right)^2 + 175\left(\dfrac{125}{7}\right) + 143\\\\\\\\h\left(\dfrac{125}{7}\right) = -1562.5 + 3125+ 143\\\\\\\\h\left(\dfrac{125}{7}\right) =1705.5$

Therefore, the rocket peaks at 1,705.5 meters above sea-level.