Hi...I believe your problem is solved but sorry for the long post:
Just to add to this, you are normally better off using the more general relationship,
<br />
y = y_{0} + (V_{0}sin\theta)t - \frac{1}{2}gt^{2}<br />
where y_{0} is the initial y-coordinate of the body (it does not necessarily start from y = 0), V_{0} is the magnitude of the projection velocity, theta is the angle made by the direction of projection with the horizontal and of course t and g are time and acceleration due to gravity.
As you have to find the maximum height, which would be given by
<br />
y_{max} = \frac{V_{0}sin^{2}\theta}{g}<br />
if the projectile were projected from the origin (x = 0, y = 0).
Now since we're projecting it from some a point whose y coordinate is y_{0}, the relation for y_{max} changes as if the axes have been translated to the point of projection. So, now the correct relation is
<br />
y_{max} = y_{0} + \frac{V_{0}sin^{2}\theta}{g}<br />
This is the equation you need for your problem. You can verify it rigourously by using the first equation above and taking its time derivative. This yields,
<br />
\frac{dy}{dt} = V_{0}sin\theta - gt<br />
Setting this equal to zero you get the time at which the maximum height is reached,
<br />
t_{ymax} = \frac{V_{0}sin\theta}{g}<br />
Now put this value of t_{ymax} into the first equation and you will get the abovementioned formula for y_{max} in the general case where the particle/projectile is projected from a point whose y-coordinate is y_{0}. In fact, if you have studied parabolas in mathematics, you will be able to associate the characteristics of parabolas with the trajectory of projectiles (which is parabolic as you already know). This helps in quickly transforming the equations.
Finally, you should keep in mind the following equations for the general case of projection from the point (x0, y0):
<br />
x = x_{0} + (V_{0}cos\theta)t<br />
<br />
y = y_{0} + (V_{0}sin\theta)t - \frac{1}{2}gt^{2}<br />
<br />
V_{x} = V_{0}cos\theta <br />
<br />
V_{y} = (V_{0}sin\theta)t - gt<br />
and the equation of trajectory:
<br />
y = y_{0} + (x-x_{0})tan\theta - \frac{1}{2}\frac{g(x-x_{0})^{2}}{V_{0}^2}sec^2\theta<br />
Hope that helps...
Cheers
Vivek
