Any help figuring this out?
How high will a football that is thrown up into the air at 48 feet per second go if it starts its flight at a height 6 feet above the ground?
I assume, given the lack of any other information, that the football is thrown straight up into the air.
So the equation for the height of the football is:
$\displaystyle y(t) = y_0 + v_0t + \frac{1}{2}at^2$
Calling the ground the origin and upward positive we get
$\displaystyle y(t) = 6 + 48t - \frac{1}{2}32t^2$
$\displaystyle y(t) = 6 + 48t - 16t^2$
Now find the maximum height.
-Dan