After how many seconds with the ball be 200 feet high?

Question

The full question:

 

h = -16t² - 80t + 1000

 

The equation above expresses the approximate height h, in feet, of an object t seconds after it is thrown vertically downward from a building 1000 feet high with an initial velocity of 80 feet per second. After how many seconds with the ball be 200 feet high?

 

I tried, but this was confusing. Please help. Thank you!

 

Answer 1

Set h=200.

 

 

200 = -16t² - 80t + 1000

 

 

Subtract 200 on both sides of the equation.

 

0 = -16t² - 80t + 800

 

 

Factor out the right side of the equation. 

 

0 = -16(t² + 5t - 50)

 

 

Set term in parenthesis equal to zero.

 

t² + 5t - 50 = 0

 

 

We can factor this using FOIL.

 

(t - 5)(t + 10) = 0

 

 

Set the factors equal to zero.

 

t - 5 = 0             and          t + 10 = 0

 

t = 5                   and         t = -10

 

 

Since we cannot have a negative value for time, we accept t=5.

 

 

It takes 5 seconds.