Question

An object was thrown upward from the top of an 80ft tower. The height h of the object after t seconds is represented by the quadratic equation h=-16t^2 + 64t + 80. After how many seconds will the object hit the ground?

Answer 1

The problem gives you:
h=-16t^2+64t+80
where
h is height in feet
t is time in secs

If it hits the ground, h = 0 so
set h=0 and solve for t:
h=-16t^2+64t+80
0=-16t^2+64t+80
0=-4t^2+16t+5
Solving using the quadratic equation yields:
t = {-0.29, 4.29}
We can throw out the negative solution leaving
t = 4.29 seconds

Answer 2

After 5 seconds object will hit the ground.

Explanation:

An object was thrown upwards form an 80 feet tower.

The height h after t seconds is represented by the equation h = -16t²+ 64t + 80

We have to find the time taken by the object to hit the ground.

At h = 0, object will hit the ground.

Now we solve the equation for h = 0

-16t² + 64t + 80 = 0

-16(t² - 4t - 5) = 0

t² - 4t - 5 = 0

t² - 5t + t - 5 = 0

t(t - 5) + 1(t - 5) = 0

(t - 5)(t + 1) = 0

t = 5, (-1)

Since time can not be negative so t = (-1) can not be the answer.

So t = 5 seconds will be the answer.