An object is launched off the top of a building, and follows a path that can be modelled using the function h(t) = –5t2 + 50t + 55, with t in seconds and h in metres. When does it reach its maximum height?

a) 0 s
b) 1
c) 5 s
d) 10 s

You'll have to use another equation to get this I think...

Use v = u + at

You need to find t.

At maximum height, v = 0.

The initial speed is obtained in the equation you were given.

h(t) = 55 + 50t + 5t^2

This is in the form:

s_t = s_o + ut + \frac12 at^2

Get the values of u and a from this equation and use them in the equation I gave you to find the time.

Post what you get! :)

Actually,you can get the max height by just finding the vertex of the parabola.

I am going to assume the equation is h=-5t^{2}+50t+55

The vertex is found by t=\frac{-b}{2a}

\frac{-50}{2(-5)}=5

The max height can be found by h=c-\frac{b^{2}}{4a}

h=55-\frac{50^{2}}{5(-5)}=155