Quadric Equations

Question

A waterfall has a height of 1100 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 16 feet per second. The height of the pebble h in feet after t seconds is given by the equation h=-16t^2 + 16t + 1100. How long after the pebble is thrown will it hit the ground?

Answer 1

It's important to realize that the height of the waterfall AND the initial velocity are both built into the equation you were given:

h = -16t² + 16t + 1100

Also, it's important to realize that the ground is at h = 0. So we just make the "h" in the equation zero and solve for "t":

0 = -16t² + 16t + 1100 [Divide both sides by 4 to simplify the equation a little:

0 = -4t² + 4t + 275

Now use the quadratic formula

For 0 = at² + bt + c, then t = (-b ± √(b² - 4ac))/(2a) so a = -4, b = 4, and c = 275 so:

t = (-b ± √(b² - 4ac))/(2a)

t = (-4 ± √(4² - 4(-4)(275))/(2(-4))

t = (-4 ± √(16 + 4400))/-8

t = (-4 ± √4416)/-8

t = (-4 ± 66.453)/-8

t = (-4 + 66.453)/-8 and t = (-4 - 66.453)/-8

t = -7.807 seconds and t = 8.807 seconds

Obviously the t= -7.807 seconds is not an answer that makes sense because it is backwards in time. So the answer is t = 8.807 seconds