[Math] Show the maximum height of particle

projectile motion

I have the question "A particle is projected so that the horizontal and vertical components of its initial velocity are U and V respectively. Show that the maximum height of the particle is:

V^2 / 2g
"

Here is my attempt however I do not get the correct result what have I done wrong ?

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Best Answer

At maximum height, the vertical velocity is zero. We also know that in simple kinematic situations with constant acceleration:

$$V_f^2 = V_0^2 + 2ax$$

And for our situation, $V_f = 0$, $V_0 = v$, $a = -g$, and $x = h$.

So we have

$$0^2 = v^2 - 2gh$$ Solve for $h$ to get

$$h = \frac{v^2}{2g}$$

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