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:
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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.
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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).
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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.
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