Answer:
hmax=81ft
Explanation:
Maximum height of the object is the highest vertical position along its trajectory.
The vertical velocity is equal to 0 (Vy = 0)
[tex]0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\[/tex]
we isolate th (needed to reach the maximum height hmax)
[tex]th = \frac{v_{0}*sin(\alpha)}{g}[/tex]
The formula describing vertical distance is:
[tex]y = Vy * t-g* t^{2} / 2[/tex]
So, given y = hmax and t = th, we can join those two equations together:
[tex]hmax = Vy* th-g*th^{2}/2[/tex]
[tex]hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)[/tex]
if we launch a projectile from some initial height h all you need to do is add this initial elevation
[tex]hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)[/tex]
[tex]hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft[/tex]
