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? Explain how you found your answer.

Answer 1

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.

Explanation:

This is the exact answer on edg 2020

Answer 2

The object hits the ground 5 seconds after being launched.

Explanation:

The height of the object in t seconds after being launched is given by the following equation:

`h(t) = 80t - 16t^(2)`

When will the object hit the ground after it is launched?

This is t for which h(t) = 0.

So

`80t - 16t^(2) = 0`

`16t(5 - t) = 0`

Then

`16t = 0`

`t = 0`

This is the launch point

`5 - t = 0`

`t = 5`

So

The object hits the ground 5 seconds after being launched.