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Step-by-Step Solution
1. Understanding the Problem
A rubber ball is dropped from a height $h = 4.9\,\text{m}$ onto a horizontal plate. The collision with the plate is perfectly elastic and very short in duration. We need to determine the velocity as a function of time and the height as a function of time, and understand their graphical representations.
2. Downward Motion
When the ball is dropped from rest, its initial velocity is zero and it accelerates downward under gravity with acceleration $g$.
Velocity as a function of time (downward):
Since the ball starts from rest, $u = 0$. The velocity after time $t$ is given by
$v = -\,g t,$
where the negative sign indicates downward direction (taking upward as positive).
Displacement (height) as a function of time (downward):
Using $y - y_{0} = ut + \tfrac{1}{2} a t^2,$ and letting $y_0 = h$ and $u=0$, we have:
$$
y - h = 0 \cdot t + \frac{1}{2}(-g) t^2 \quad\Longrightarrow\quad y = h - \frac{1}{2} g t^2.
$$
Thus, at $t=0,$ $y = h$, and as time increases, $y$ decreases, forming a downward-opening parabola in the $y$–$t$ plot.
3. Collision and Upward Motion
The collision with the horizontal plate is perfectly elastic. This means the speed just after collision has the same magnitude as just before collision, but the direction reverses (now positive, i.e., upward).
Velocity just after collision:
Right before collision, $v = -\,gT,$ where $T$ is the time taken to reach the plate. Right after collision, the velocity becomes $+\,gT$ (same magnitude, opposite direction).
Velocity as a function of time (upward):
If $u$ is the velocity just after collision, the velocity after time $t$ (measured from the collision instant) is
$$
v = u - g t.
$$
Since $u = gT$ and $g$ remains in the same direction (downward), the plot of $v$ versus $t$ continues as a straight line with negative slope.
Height as a function of time (upward):
From the collision time onward, using
$$
y = ut - \frac{1}{2} g t^2,
$$
where $u = gT$, the motion is again a parabola in $y$–$t$ with the vertex at the collision moment and opening downward.
4. Graphical Representation
The correct option shows:
A straight-line $v$–$t$ graph with negative slope both during downward and upward motion, but velocity sign flips at the collision.
A downward-opening parabola for $y$–$t$ on both sides of the collision (first decreasing from $h$ to $0$, then increasing back from $0$). The overall effect combines two such parabolas, joined at the collision point.
The option that best matches these characteristics is the one with continuous linear $v$–$t$ segments changing sign at the collision and the parabolic $y$–$t$ path.
5. Final Answer
The correct graphs are represented by the provided option:
