What is the velocity of an object at its maximum height when it is thrown upward from ground?

Maximum height of the object is the highest vertical position along its trajectory. The object is flying upwards before reaching the highest point - and it's falling after that point. It means that at the highest point of projectile motion, the vertical velocity is equal to 0 (Vy = 0).

0 = Vy – g * t = V₀ * sin(α) – g * th

From that equation we can find the time th needed to reach the maximum height hmax:

th = V₀ * sin(α) / g

The formula describing vertical distance is:

y = Vy * t – g * t² / 2

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

hmax = Vy * th – g * th² / 2

hmax = V₀² * sin(α)² / g – g * (V₀ * sin(α) / g)² / 2

hmax = V₀² * sin(α)² / (2 * g)

And what if we launch a projectile from some initial height h? No worries! Apparently, the calculations are a piece of cake - all you need to do is add this initial elevation!

hmax = h + V₀² * sin(α)² / (2 * g)

Let's discuss some special cases with changing angle of launch:

if α = 90°, then the formula simplifies to:

hmax = h + V₀² / (2 * g) and the time of flight is the longest.

If, additionally, Vy = 0, then it's the case of free fall. Also, you may want to have a look at our even more accurate equivalent - the free fall with air resistance calculator.
if α = 45°, then the equation may be written as:

hmax = h + V₀² / (4 * g) and in that case, the range is maximal if launching from the ground (h = 0).
if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion. As sine of 0° is 0, then the second part of the equation disappears, and we obtain :

hmax = h - initial height from which we're launching the object is the maximum height in projectile motion.