Question

Question #8b7fa

Answer 1

The object is at a height of `3/4 * h` from ground level.

Explanation:

So, you know that your object is being released from a height `h` and that it reaches the ground in a time `t`.

This means that you can write

`h = underbrace(v_0)_(color(blue)(=0)) * t + 1/2 * g * t^2`

Now, let's assume that we can stop this object's motion at `t/2` and see how far it fell from its initial position `h`. This distance, let's say `h_1`, will be

`h_1 = underbrace(v_0)_(color(blue)(=0)) * t/2 + 1/2 * g * (t/2)^2`

`h_1 = 1/2 * g * t^2/4`

Notice that `h_1` is actually equal to

`h_1 = underbrace(1/2 * g * t^2)_(color(blue)(=h)) * 1/4 = 1/4 * h`

Well, if the object travelled `h/4` in `t/2`, then its distance from the ground is

`h_"from ground" = h - h_1`

`h_"from ground" = h - h * 1/4 = color(green)(3/4 * h)`