Question

A rocket is fired with an initial vertical velocity of 40 m/s from a launch pad 45 m high, and its height is given by the formula -5t^2+40t+45 where h is in meters and t is in seconds. How high does the rocket go? How much time is the rocket in flight?

Answer 1

Answer:

1) 125 meters

2) For 9 seconds.

Step-by-step explanation:

The rocket's height is given by the formula $-5t^2+40t+45$.

Notice that this is a quadratic.

Part 1)

Since this is a quadratic, the highest height the rocket goes will be the vertex of our quadratic. Remember that the vertex of a quadratic in standard form is:

$(-\frac{b}{2a}, f(-\frac{b}{2a}))$

So, let's identify our coefficients. The standard quadratic form is:

$ax^2+bx+c$

Therefore, in this case, our a is -5, b is 40, and c is 45.

So, let's find the x-coordinate of our vertex. Substitute 40 for b and -5 for a. This yields:

$x=\frac{-(40)}{2(-5)}$

Multiply and reduce. So:

$x=-40/-10=4$

Now, substitute 4 for t to find the height. So:

$-5(4)^2+40(4)+45$

Evaluate:

$=-5(16)+40(4)+45$

Multiply and add:

$=-80+160+45=125\text{ meters}$

Therefore, the highest the rocket goes up is 125 meters.

Part 2)

To find out for how much time the rocket is in the air, we can think about after how many seconds after launch will the rocket land.

If the rocket lands, the height h will be 0. So, we can set our expression equal to 0 and solve for t:

$-5t^2+40t+45=0$

First, let's divide everything by -5. This yields:

$t^2-8t-9=0$

We can factor:

$(t+1)(t-9)=0$

Zero Product Property:

$t+1=0\text{ or } t-9=0$

Solve for t:

$t=-1\text{ or } t=9$

In this case, since t is our time in seconds, -1 seconds does not make sense. So, we can remove -1 from our solution set.

Therefore, 9 seconds after launch, the rocket will touch the ground.

Therefore, the rocket was in the air for 9 seconds.

And we're done!