An object is launched from the ground. The object’s height, in feet, can be described by the quadratic function h(t) = 80t – 16t
2, 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.
Would you like me to help walk you through this with you figuring out the work? Or would you prefer just an answer in math form with me explaining it along the way?
<span>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.
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.
