A projectile is launched from a cannon that sits 4 feet off the ground at an initial velocity of 120 ft/sec. Find the position function and how long is it before the projectile hits the ground?
I got -16t^2 +120t+4 for my position function, but I'm not sure how to find the other part?
2 answers
Also, how do I find the maximum height?
let h(t) be the position function
h(t) = - 16t^2 + 120t + 4 . you had that.
when it hits the ground, h(t) = 0
-16t^2 + 120t + 4 = 0
4t^2 - 30t -1 = 0
solve with quadratic formula
t = (30 ± √916)/8 = 7.533 or a negative value of t
it will hit the ground after appr 7.5 seconds
h(t) is a parabola opening downwards, so it has a maximum at its vertex.
I don't know what method you have learned to find the vertex, a common method is to complete the square.
OR
the x of the vertex is -b/(2a) = -120/-32 = 15/4 or 3.75
h(15/4) = -16(225/16) + 120(15/4+ + 4
= 229 ft
h(t) = - 16t^2 + 120t + 4 . you had that.
when it hits the ground, h(t) = 0
-16t^2 + 120t + 4 = 0
4t^2 - 30t -1 = 0
solve with quadratic formula
t = (30 ± √916)/8 = 7.533 or a negative value of t
it will hit the ground after appr 7.5 seconds
h(t) is a parabola opening downwards, so it has a maximum at its vertex.
I don't know what method you have learned to find the vertex, a common method is to complete the square.
OR
the x of the vertex is -b/(2a) = -120/-32 = 15/4 or 3.75
h(15/4) = -16(225/16) + 120(15/4+ + 4
= 229 ft
2 answers