Question #8b7fa

1 Answer
Sep 6, 2015

Answer:

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)#