A tennis ball is released from a height h and | vedclass

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

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(A) The motion of the ball under gravity is governed by the equation $v^2 = u^2 + 2as$. For a ball falling from rest at height $h$, the velocity $v$ at any height $y$ is given by $v^2 = 2g(y - h)$ (taking downward as positive direction for velocity during fall). $1$. 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 heights $h$ and $h/2$), the slope $\frac{dv}{dy} 	o \infty$. This means the graph must be vertical at the points where $v=0$. $6$. Graph $A$ correctly represents this behavior, showing the parabolic nature of the $v-y$ relationship and the vertical tangents 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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