Explanation

(b) When a ball is dropped on a floor then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = \frac{1}{2}\ gt^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
So, the graph between $yand\ t$ is a parabola. Here $y$ decrease as time decrease. When the ball is bounces back then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = ut - \frac{1}{2}gt^{2}\ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Eq. (ii) is also the form of a general equation of parabola so the graph between $y\ and\ t$ will be a parabola. Here $y$ increases when the time increases. Hence, the required graph between $y\ and\ t$is shown in the figure.
[image:3_image204.emf]