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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?

Sagot :

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 [tex]f(t)=-5t^2+20t+60[/tex] 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.

[tex]-5t^2+20t+60=0[/tex]

Upon dividing our equation by -5 we will get,

[tex]t^2-4t-12=0[/tex]

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

[tex]t^2-6t+2t-12=0[/tex]

[tex]t(t-6)+2(t-6)=0[/tex]

[tex](t-6)(t+2)=0[/tex]

[tex](t-6)=0\text{ or }(t+2)=0[/tex]

[tex]t=6\text{ or }t=-2[/tex]

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

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