Question

The function t)=-5t^2+20t+60" align="absmiddle" class="latex-formula"> models the approximate height of an object t seconds after it is launched. How many seconds does it take the object to hit the ground?
(Use without A CALCULATOR)

Answer 1

Basically when the object hits the ground, f(t) is equal zero. So all you need to do is to solve the quadratic equation:
-5t^2 + 20t + 60 = 0

D = b^2 - 4ac = 20^2 - 4 × (-5) × 60 = 400 + 1200 = 1600

t1 = (-b + sqrt(D)) / 2a = (-20 + 40) / (-10) = -2
t2 = (-b - sqrt(D)) / 2a = (-20 - 40) / (-10) = 6

Since time can't be negative, our solution is t2.

So the object will hit the ground in 6 seconds.