Question in Description

Question

A waterfall has a height of 1400 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 12 feet per second. The height , h, of the pebble after t seconds is given by the equation h=-16t^2+12t+1400. How long after the pebble is thrown will it hit the ground?

Answer 1

Hi Rebekah,

 

This question is tricky because it gives you a bunch of information that you don't need. Let's focus on the height equation to figure it out.

 

h = -16t² +12t +1400

 

This equation tells us the height of the pebble at any time. What is the height when it is on the ground? Think logically, and the height is 0, right?

 

So, make h=0 and we get

 

0 = -16t² +12t +1400

 

How about that! We have a quadratic equation in standard form. We can now use a few different methods to solve for 't' - factoring, the quadratic formula, or graphing.

 

Factoring gets kind of ugly, but let's see what happens when we do that. If you noticed, each coefficient is divisible by 4. Factoring out the negative sign on the x² term is useful, as well. We get...

 

0 = -4 * (4t² - 3t - 350)

 

Unfortunately, that doesn't factor easily.

 

As an alternate route, use the quadratic equation. This also gets messy very quickly.

 

For a last resort, graph it on your calculator (or use desmos.com/calculator) and see when the function is equal to 0. You should find that the function is equal to 0 when t = -8.987 or 9.737.

 

The negative value of time makes no sense, as clocks always tick forward. So, the solution is 9.7 seconds.

 

Note: If you use the desmos online calculator, you need to replace the t's in your given equation with x's.

 

I hope this helps!