Question

A body undergoes a free fall. Taking downward direction as +ve $Y$ direction and $Y_o$ is the initial position, $Y$ coordinate at any time $t$ is given by

• A. $Y_o-Y=\frac{1}{2}g{t}^2$
• B. $Y_o=\frac{1}{2}g{t}^2$
• C. $Y_o-Y=-\frac{1}{2}g{t}^2$
• D. $Y_o+Y=\frac{1}{2}g{t}^2$

Answer 1

The correct option is C $Y_o-Y=-\frac{1}{2}g{t}^2$

Initial position is $Y_o$ and downward is +ve $Y$ direction so gravity $\left(g\right)$ is acting in positive direction.
Displacement, $s=ut+\frac{1}{2}a{t}^2,u=0anda=g$
We get, $s=0+\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$
So, let the final position be $Y=Y_o+s=Y_o+\frac{1}{2}g{t}^2$
$\therefore Y_o-Y=-\frac{1}{2}g{t}^2$