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,
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):
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.
