# Hot-Air Balloon

• January 15th 2007, 12:29 PM
symmetry
Hot-Air Balloon
A hot-air balloon, headed due east at an average speed of 15 mph and at a constant altitude of 100 feet, passes over an intersection. Find an expression for its distance d (in feet) from the intersection t seconds later.

Do I use D = rt?

If so, then D = 100 feet, right?

Then r = 15 mph, right?

I got:

100 = 15(t)

100/15 = t

Am I right or wrong?

If wrong, what did I do wrong?
• January 15th 2007, 01:04 PM
topsquark
Quote:

Originally Posted by symmetry
A hot-air balloon, headed due east at an average speed of 15 mph and at a constant altitude of 100 feet, passes over an intersection. Find an expression for its distance d (in feet) from the intersection t seconds later.

Do I use D = rt?

If so, then D = 100 feet, right?

Then r = 15 mph, right?

I got:

100 = 15(t)

100/15 = t

Am I right or wrong?

If wrong, what did I do wrong?

$x = rt$ is the balloon's horizontal distance from the intersection at t seconds.

So $x = 15t$.

BUT t is in seconds and the 15 is mi/h, so we need to change the unit. In addition we will want x to be in feet, so we need to change miles to feet. There are 3600 s in 1 h, and there are 5280 feet in 1 mile:
$\frac{15 \, mi}{1 \, hr} \cdot \frac{1 \, hr}{3600 \, s} \cdot \frac{5280 \, ft}{1 \, mi} = 22 ft/s$

So $x = 22t$ in the units we want.

Now, the balloon is at an altitude of 100 ft, so by the Pythagorean theorem the balloon is a distance of $\sqrt{x^2 + (100)^2}$ feet from the intersection.
$d = \sqrt{(22t)^2 + (100)^2} = \sqrt{484t^2 + 10000}$ feet.

-Dan
• January 15th 2007, 03:56 PM
symmetry
ok
I thank you for going the extra mile with this question.

I don't think the question asked for an answer but it's good to know that an expression and answer were displayed by you here.

Thanks.