waterfall question

Question

A waterfall has a height of 1500 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 16 feet per second. The height,h, of the pebble after t seconds is given by the equation h=-16t^2+16t+1500. How long after the pebble is thrown will it hit the ground?

Answer 1

Hi James,

 

What will the height of the pebble be when it hits the ground? It will be at h = 0.

 

h = -16t² + 16t + 1500

0 = -16t² + 16t + 1500

 

All 3 terms have a common factor of 4, so divide both sides by 4 to make life easier.

0 = -4t² + 4t + 375

 

You can either try to solve this by factoring, or just use the quadratic formula. Since factoring 375 isn't fun, I recommend the quadratic formula.

 

t = [-b ± √(b² - 4ac)] / 2a

 

From your equation,

a = -4

b = 4

c = 375

 

t = [-4 ± √(4² - 4(-4)(375))] / 2(-4)

t = [-4 ± √(6016)] / -8

 

You get two answers: one positive, and one negative. You can ignore the negative result, and the positive number is your answer. I get a value around 10 s.