An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t^2 + 64t + 80?

Question

campbell_st · Answer

you need to find the time that the max height occurs. 

have you done calculus...?

anonymous · Answer

no, just algebra 2

campbell_st · Answer

ok... that helps.... there are a few methods

I'd recommend you find the line of symmetry

for a parabola 

$$at^2 +bt + c = 0$$

the line of symmetry is 

$$t = \frac{-b}{2a}$$

you have a = -16 and b = 64

can you substitute them to find t, the time the max height occurs..?

campbell_st · Answer

an alternate method is to divide every term by -16 and then put the parabola into the vertex form

y = (x - h)^2 + k

anonymous · Answer

what's x, h and k?

campbell_st · Answer

(h, k) is the vertex, it can be used as an alternate method. 

my advice is to use the 1st method I posted. Which is using the line of symmetry.

anonymous · Answer

ohhh okay

anonymous · Answer

but that leaves me with -2??? @campbell_st

anonymous · Answer

how is that correct? did I do something wrong? @Campbell_st

campbell_st · Answer

it leaves you with 

t = 64/(-2x-16)

so at t = 2 the pumpkin reaches its maximum height.

so then find 

h(2) = -16x(2)^2 + 64 x 2 + 80

this will give the max height.