SOLUTION: An object is launched straight up into the air at an initial velocity of 64 feet per second. It is launched from an initial height of 6 feet off the ground. Find all times t that t
Question 1152646: An object is launched straight up into the air at an initial velocity of 64 feet per second. It is launched from an initial height of 6 feet off the ground. Find all times t that the object is at a height of 54 feet off the ground.
You can put this solution on YOUR website! An object is launched straight up into the air at an initial velocity of 64 feet per second.
It is launched from an initial height of 6 feet off the ground.
Find all times t that the object is at a height of 54 feet off the ground.
:
Using the formula h = -16t^2 + vt + c, where
h = the height above ground at t time
t = time since initially launched
v = initial velocity
c = height above ground at launch
:
h = 54
-16t^2 + 64t + 6 = 54
-16t^2 + 64t + 6 - 54 = 0
-16t^2 + 64t - 48 = 0
simplify, divide by -16
t^2 - 4t + 3 = 0
factors to
(t-3)(t-1) = 0
two solutions
t = 1 sec at 54 ft on the way up
and
t = 3 sec at 54 ft on the way down
