# Proof of a formula relating launch velocity of projectile to times of ascent and descent

I came across this formula in my textbook stating that when a body is projected vertically upwards and the time corresponding to height $H$ while ascending and descending are $t_1$ and $t_2$ respectively, when the velocity of projection is $\frac{g(t_1+t_2)}{2}$ where $g$ is acceleration due to gravity.

Can anyone please help me in deriving this formula? I know that in case of vertical motion upwards the initial velocity will be gt1. But I don't know how to proceed further. Please help.

• You should add some details in the title. It will be more likely to interest readers when you know better the content. – Riccardo Buscicchio Apr 29 '17 at 8:17
• By the way, your statement about the time $gt_1$ being the initial velocity holds when $H$ is the maximum height, not a general one. Try and derive this result, for better understanding. – Riccardo Buscicchio Apr 29 '17 at 8:25

The height $h(t)$ of the body as a function of time is $$h(t)=-\frac{1}{2}gt^2+vt$$ with $v$ the initial speed. At $t_1$ and $t_2$, the body will be at the same height. It means that $$h(t_1)=h(t_2)$$ or, equivalently $$-\frac{1}{2}gt_1^2+vt_1=-\frac{1}{2}gt_2^2+vt_2 \\\Downarrow \\v(t_1-t_2)=\frac{1}{2}g(t_1^2-t_2^2) \\\Downarrow \\ v=\frac{g(t_1^2-t_2^2)}{2(t_1-t_2)}=g\frac{(t_1+t_2)}{2}$$
Average Velocity between two points at the different height: $V_(av)$ = $\frac{V_1 +V_2}{2}$ =$\frac{gt_1 +gt_2}{2}$ =$\frac{g(t_1+t_2)}{2}$