# Height of a falling ball

I'm confused about this formula $H = \frac {g}{2}\cdot t^2$

Imagine the following situation. A ball falls from a wall of $125$m and the fall lasts $5$s. At the height of $80$m the time is $4$s, at the height of $45m$ the time is $3$s, since $80+45 = 125$, the time required to fall from a height of $125m$ is $4+3=7$, but by the formula the time would be $5$ and not $7$, I'm confused about this.

• You can't simply add up square roots. – Gert Dec 6 '17 at 17:32

The formula you have given $$H=\frac{gt^2}{2}$$ is correct in the case where the initial velocity is $0$. This means that the time you have calculated is longer, as for the second part you used this equation which incorrectly assumes that at that point the initial speed would be $0$. The full equation should be $$H=ut+\frac{gt^2}{2}$$Hope this helps :)
In the case that throughout the question the initial velocity is zero (i.e. the object is dropped from a height $H$ ):
As the equation requires the time to be squared, the addition of the two times would have to be done like this: $$t_1^2+t_2^2=t_{total}^2$$ So $$4^2+3^2=25=5^2$$