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

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h(t) = 80t – 16t^2
When you are on the ground we have zero height, h(t) = 0. This is saying that at some amount of time we will call t, the object will hit the ground. Now we solve for t to get that time 
so we sub h(t) for 0.
0=80t-16t^2 move the -16t^2 to the other side
 16t^2=80t   divide both sides by t
16t=80   divide both sides by 16
t=5
Now... it is important to realize that there are going to be 2 answers to this equation as you have a t^2. The other answer is 0. Obviously, unless the object started on the ground, the time to hit the ground is not 0 and that answer is ruled out.

Answer 2

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.