Please help! If the path of a projectile is given by h=-4t^2+256t, determine its maximum height. How would I go about solving this?
Options:
a. 8192
b. 4096
c. 3072
d. 16,384

Question

Please help! If the path of a projectile is given by h=-4t^2+256t, determine its maximum height. How would I go about solving this?
Options:
a. 8192
b. 4096
c. 3072
d. 16,384

anonymous · Answer

max is a vertex, namely 
$$-\frac{b}{2a}=-\frac{256}{-4}=16$$ then replace x by 16

anonymous · Answer

are you sure it isn't 
$$h=-16t^2+256t$$?

anonymous · Answer

what is h and t ?

anonymous · Answer

@satellite, Yes I'm sure it isn't ... The problem asks for h=-4t^2+256t. h stands for height but I don't know about t.

anonymous · Answer

I got C. Can someone verify?

ash2326 · Answer

We have 
$$h= -4t^2+256 t$$
vertex 
$$\frac{b}{2a}=-\frac{256}{8}=32$$
this is the value of t at which height is max
$$h= -4t^2+256 t$$
t=32
$$h=-4 \times 32^2+256\times 32= 32(256-128)=4096 meters$$

anonymous · Answer

Oh that makes sense. Thank you for the explanation!