Question

A model rocket is launched from the top of a building. The height(in meters) of the rocket above the ground is given by h(t)=-6t^(2)+24t+14, where t is the time (in seconds) since the rocket was launched. What is the rocket's maximum height?

Answer 1

Answer:

Step-by-step explanation:

I don't know if you are a calculus student, but since calculus is the easiest way to solve this, that's what I used.  The position equation is

$s(t)=-6t^2+24t+14$

and the velocity function, the first derivative, is

$v(t)=-12t+24$

From Physics, you should know that the velocity of an object is 0 at its highest point.  That means that if we sub in a 0 for the velocity and solve for t, we find the time at which the object is at its highest point.  That is:

$0=-12t+24$ and

$0=-12(t-2)$

By the Zero Product Property,

-12 definitely does not equal 0, so that means that

t - 2 = 0 and

t = 2 seconds.

So at 2 seconds, the object is at its highest point.  Go back to the position function now and sub in a 2 for t to find the height of the object at 2 seconds.

$s(2)=-6(2)^2+24(2)+14$ so

s(2) = 38 meters