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mcnman31 · Nov 2, 2010, 08:55 AM

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

Unknown008 · Nov 2, 2010, 10:34 PM

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! :)

galactus · Nov 3, 2010, 04:59 AM

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 $h=c-\frac{b^{2}}{4a}$

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