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

Answer:

The object hits the ground 5 seconds after being launched.

Step-by-step 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.

Answer 2

Answer:

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.

Step-by-step explanation:

This is the exact answer on edg 2020