Here object is dropped from height "h"
so we can say its initial speed is zero
and it will accelerate downwards due to gravity
now we can say it will take time T to hit the ground
now we can use
[tex]h = v_i*t + \frac{1}{2}gt^2[/tex]
[tex]h = 0 + \frac{1}{2}*gT^2[/tex]
now it is given that it will take 1 second to drop h/2 height to strike the ground
so here we have can say that in "T - 1" s it will cover the h/2 distance from start
[tex]h/2 = 0 + \frac{1}{2}g(T-1)^2[/tex]
now we can say
[tex]h = g(T-1)^2[/tex]
from above two equations we have
[tex]\frac{1}{2}gT^2 = g(T-1)^2[/tex]
[tex]T^2 = 2(T^2 + 1 -2T)[/tex]
[tex]T^2 - 4T + 2 = 0[/tex]
[tex]T = 3.41 s[/tex]
now we can find total height of the drop by first equation
[tex]h = \frac{1}{2}gT^2[/tex]
[tex]h = \frac{1}{2}*9.81*3.41^2[/tex]
[tex]h = 57.2 m[/tex]
