Abaseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/sec^2) find the maximum height

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)

we isolate th (needed to reach the maximum height hmax)

The formula describing vertical distance is:

So, given y = hmax and t = th, we can join those two equations together:

if we launch a projectile from some initial height h all you need to do is add this initial elevation

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at right angles to each other