Expert: Sherry Wallin - 4/28/2009

Question
One rocket has an initial velocity of 48 feet per second. The other has an initial velocity of 64 feet per second.
Question 1: What is the maximum height the rockets will reach and at what time in seconds did they reach the max height?
Question 2: Find the time in seconds when the rockets are at 0 feet. Why does this happen twice?
Your help would be greatly appreciated. thanks

Answer
Hi Carrie~
    The equation you sent me is the standard quadratic equation for describing what an object on earth travels, ie., a parabolic path. Here in your first question since the initial velocity is 48ft/s and the initial height is presumed to be 0 our equation is h(t) = -16t^2 + 48t
You want to know the maximum height and this is simply the vertex of the parabola and you know that -b/2a is the x coordinate of the vertex (which is the time) where a = -16 and b = 48, so -b/2a is -48/(2*(-16)) = 3/2 which is the time in seconds for the rocket to reach it's maximum height. The y coordinate is f(3/2) = -16(3/2)^2 + 48(3/2) = -16(9/4)+72 = -36+72 = 36ft. The 2nd rocket which has an initial velocity of 64ft/s has an equation of h(t) = -16t^2 + 64t, so it's vertex is -64/[2(-16)] = 2 sec and f(2)= -16(2^2) + 64(2) = -64 + 128 = 64ft. So it takes 2 sec to reach the maximum height and that height is 64ft.

With regards to the second question: h(0) = 0 + 0 = 0 (in both cases) this happens twice because if you factor each equation you get two values for the quadratic

h(t) = 0 -> 0 = -16t^2 + 48t = -16t(t-3) so -16t = 0 -> t = 0 or t-3 = 0 which implies t = 3secs and likewise with the other rocket.
-16t^2+64 = 0 -> -16t(t-4) = 0 -> t = 0 or t = 4 sec

Math Prof