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