question: ⇒ an object is launched from the ground. the object’s height, in feet, can be described by the quadratic function h(t) = 80t – 16t2, where t is the time, in seconds, since the object was launched. when will the object hit the ground after it is launched?
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you!

it would take 5 seconds for the object to hit the ground after launched.

1st step: (turn the quadratic function to a quadratic equation by setting the equal to zero): 0 = -16t2 + 80t + 0

2nd step: t = 0 & t = 5

(t = 0 is the time of launch. t = 5 represents the 5 seconds it took to hit the ground after the object was launched.)

The object will hit the ground after 5 seconds. You can rewrite the quadratic function as a quadratic equation set equal to zero to find the zeros of the function 0 = –16t2 + 80t + 0. You can factor or use the quadratic formula to get t = 0 and t = 5. Therefore, it is on the ground at t = 0 (time of launch) and then hits the ground at t = 5 seconds

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation: