SOLUTION: Please help me solve this problem below. An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H,

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Question 1090948: Please help me solve this problem below.
An object is launched into the air from a ledge 6 feet off the ground at initial vertical velocity of 96 feet per second. Its height H, in feet, at t second is given by the equation H = 16t^2+96t+16. Find all the times t that the object is at a height of 160 feet off the ground.

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
First you need to learn THIS:

    When you consider the problems like this one, the correct form of the equation for the height is 

    H(t) = -16t^2 + 96t + 16.


    Notice the sign "-" (minus) at the quadratic term.

    Your writing in the post was incorrect.


Second, all you need to do is to solve the equation H(t) = 0, or, which is the same,

= 0.

For it, divide both sides by 16. You will get

-t^2 + 6t + 1 = 0,    or, which is simpler (and equivalent !)

t^2 -6t -1 = 0.


 =  =  = .


Only positive root makes sense.


Answer.  t =  seconds.

Solved.



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