Quadratic Projectile Motion help?

Okay so here is the original equation:

A rocket is launched into the air from the top of a building. It's height (in meters) is given by:

h(t)= -5tē + 30t + 100, where t = time in seconds.

When does the rocket smash into the ground?

So, this is where I set the equation equal to zero.

0 = -5tē + 30t + 100

and since it couldn't be factored, I decided I needed to use the quadratic formula.

However, am I able to factor out the -5?

Then I got:

-5(tē-6t-20)

So, when I plugged the items in the parenthesis into the quadratic formula, I eventually got: (3 +/- √29)

Now, what I'm wondering is:

First of all, did I do everything correctly?

If so, what do I then do with the -5 I factored out in the beginning?

Any help would be appreciated, thank you for your time!

Re: Quadratic Projectile Motion help?

Quote:

Originally Posted by

**ZzTop0503** Okay so here is the original equation:

A rocket is launched into the air from the top of a building. It's height (in meters) is given by:

h(t)= -5tē + 30t + 100, where t = time in seconds.

When does the rocket smash into the ground?

So, this is where I set the equation equal to zero.

0 = -5tē + 30t + 100

and since it couldn't be factored, I decided I needed to use the quadratic formula.

However, am I able to factor out the -5?

Then I got:

-5(tē-6t-20)

So, when I plugged the items in the parenthesis into the quadratic formula, I eventually got: (3 +/- √29)

Now, what I'm wondering is:

First of all, did I do everything correctly?

If so, what do I then do with the -5 I factored out in the beginning?

Any help would be appreciated, thank you for your time!

It doesn't matter whether you factor out a 5 or not, you still get the same answer.

Your answer is correct. If you look at the parabola, you will notice that it is shot from 100 meters (the building is 100 meters tall) and it eventually lands after your value t...Which means it travels for roughly 8 seconds.

Re: Quadratic Projectile Motion help?

To check that it really doeesn't matter than you divided through by 5 you can plug the values of $\displaystyle 3 \pm sqrt {29}$ into the original formula and see that it works:

$\displaystyle -5(3 + \sqrt{29})^2 + 30(3 + \sqrt{29})t +100 = -5(9 + 6\sqrt{29} + 29) + (90 + 30 \sqrt{29}) + 100 $

$\displaystyle = -45 -30 \sqrt{29}-145 + 90 + 30 \sqrt{29} +100 = 0$

Remember that when you do algebra you can multiply or divide both sides of an equaton by any number (other than 0) and the equation still holds. For example, if x = 2 then the equation x-2 = 0 is true, but so are 2x-4 =0, 6x-6 = 0, 8x-8 = 0, etc.

By the way, the solution $\displaystyle 3 - \sqrt{29}$ satisies the equation, but not the premise that the rocket is launched at t=0. You can eliminate that answer as not being consistent with the problem statement, as it can be assumed that prior to being launched the rocket was stationary on top of the building.