Question

A ball is thrown straight up from the roof of an 80-foot building and it's height is modeled by the h(t) = â17t2 + 68t + 80, where h is the height in feet and t is time in seconds. How long (in sec) will it take the ball to hit the roof of the building on its way down? That is, solve h(t) = 80. (Enter an exact number.)

Answer 1

Answer:

t = 4

Step-by-step explanation:

Given that:

$h(t) = 17t^2 + 68t + 80$

When h(t) = 80, we have that:

$80 = -17t^2 + 68t + 80\\\\=> 0 = -17t^2 + 68t\\\\=> 17t^2 - 68t= 0\\\\=> 17t(t - 4) = 0\\\\=> t = 0, t = 4$

t = 0 represents when the ball has not been thrown at all, in its initial position.

t = 4 represents the time taken for the ball to hit the roof of the building on its way down.

Hence, our answer is t = 4. (t is in secs)

Answer 2

Answer:

t = 4 seconds.

it takes the ball 4 seconds to hit the roof of the building on its way down

A ball is thrown straight up from the roof of an 80-foot building and it's height is modeled by the h(t) = -17t^2 + 68t + 80, where h is the height in feet and t is time in seconds. How long (in sec) will it take the ball to hit the roof of the building on its way down? That is, solve h(t) = 80. (Enter an exact number.)

Step-by-step explanation:

Given that;

h(t) = -17t^2 + 68t + 80

At h(t) = 80

-17t^2 + 68t + 80 = 80

-17t^2 + 68t = 80 - 80

-17t^2 + 68t = 0

-17t(t-4) = 0

So,

-17t = 0 or. t-4 = 0

t = 0 or. t = 4

Since t cannot be zero, since t = 0 is the time the ball was thrown up.

t = 4 seconds.

it take the ball 4 seconds to hit the roof of the building on its way down