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(D) The motion of the ball under gravity is governed by the equation of motion $v^2 = u^2 + 2as$. For a ball falling from rest at height $h$,the veloc

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(D) The motion of the ball under gravity is governed by the equation of motion $v^2 = u^2 + 2as$. For a ball falling from rest at height $h$,the velocity $v$ at any height $y$ is given by $v^2 = 2g(h - y)$.$1$. During the downward motion from $h$ to $0$,the velocity increases from $0$ to $\sqrt{2gh}$. The relation $v = \sqrt{2g(h-y)}$ shows that $v$ is non-linear with respect to $y$.$2$. Upon impact with the floor,the velocity changes instantaneously from $-\sqrt{2gh}$ to $+\sqrt{2g(h/2)} = \sqrt{gh}$.$3$. During the upward motion from $0$ to $h/2$,the velocity decreases from $\sqrt{gh}$ to $0$ following $v^2 = 2g(h/2 - y)$.$4$. Differentiating $v^2 = 2g(h-y)$ with respect to $y$,we get $2v \frac{dv}{dy} = -2g$,which implies $\frac{dv}{dy} = -\frac{g}{v}$.$5$. As $v 	o 0$ (at maximum height),the slope $\frac{dv}{dy} 	o \infty$. This means the graph must be vertical at the points where $v=0$ (i.e.,at $h$ and $h/2$).$6$. Graph $D$ correctly represents this behavior,showing the non-linear relationship and the infinite slope at the maximum heights.

$A$ tennis ball is released from a height $h$ and after freely falling on a wooden floor,it rebounds and reaches a height $\frac{h}{2}$. The velocity versus height of the ball during its motion may be represented graphically by (graphs are drawn schematically and not to scale):

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