Answered

The function f(t) = −5t 2 + 20t + 60 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?

Verified Answer :

kayt91

f(t) = -5t^2 + 20t + 60
-5(t^2 - 4t - 12)
-5(t + 2) (t - 6)
t = -2 or t = 6

6 seconds

Answer:

6 seconds.

Step-by-step explanation:

We have been given a function formula \(f(t)=-5t^2+20t+60\) that models the approximate height of an object  t seconds after it is launched.

To find the number of seconds it will take the object to hit the ground, we need to find the zeros of our given function.

\(-5t^2+20t+60=0\)

Upon dividing our equation by -5 we will get,

\(t^2-4t-12=0\)

We will factor our given equation by splitting the middle term as:

\(t^2-6t+2t-12=0\)

\(t(t-6)+2(t-6)=0\)

\((t-6)(t+2)=0\)

\((t-6)=0\text{ or }(t+2)=0\)

\(t=6\text{ or }t=-2\)

Since time cannot be negative, therefore, it will take the object 6 seconds to hit the ground.

Related Questions