Question

An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reached by the rocket?

Answer 1

Answer:

The maximum height is 784 feet

Step-by-step explanation:

In this problem we use the kinematic equation of the height h of an object as a function of time

$h(t) = -16t ^ 2 + v_0t + h_0$

Where $v_0$ is the initial velocity and $h_0$ is the initial height.

We know that

$v_0 = 192\ \frac{ft}{sec}$

$h_0 = 208\ ft.$

Then the equation of the height is:

$h(t) = -16t ^ 2 + 192t +208$

For a quadratic function of the form $ax ^ 2 + bx + c$

where $a$

the maximum height of the function is at its vertex.

The vertice is

$x = -\frac{b}{2a}\\\\y = f(\frac{-b}{2a})$

In this case

$a = -16\\b = 192\\c = 208$

Then the vertice is:

$t = -\frac{192}{2(-16)}\\\\t = 6\ sec$

Now we calculate h (6)

$h(6) = -16(6) ^ 2 +192(6) +208\\\\h(6) = 784\ feet$

The maximum height is 784 feet