A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h/2. The velocity versus height of the ball during its motion may be represented graphically by: (graph are drawn schematically and on not to scale) (a) (b) (c) (d)

Answer: (c). Falling from h (downward, v < 0): v² = 2g(h - y), so |v| grows from 0 at y = h to √(2gh) at y = 0 along a curve. At the floor the velocity reverses: it jumps to a positive value √(gh) at y = 0. Rising: v² = 2g(h/2 - y), so v falls to 0 at y = h/2. Curve from h to the negative v -axis, then from the positive v -axis to h/2: answer (c).

A Tennis ball is released from a height h and after freely falling on… | JEE Main 2020 (4 Sep, Shift 1)

Q43JEE Main 2020 · 4 SEP · SHIFT 1Medium

Motion Under Gravity

A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height $\frac{h}{2}$ . The velocity versus height of the ball during its motion may be represented graphically by:

(graph are drawn schematically and on not to scale)

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Answer: (c)

Falling from $h$ (downward, $v < 0$ ): $v^2 = 2g(h - y)$ , so $|v|$ grows from 0 at $y = h$ to $\sqrt{2gh}$ at $y = 0$ along a curve.
At the floor the velocity reverses: it jumps to a positive value $\sqrt{gh}$ at $y = 0$ .
Rising: $v^2 = 2g(\frac{h}{2} - y)$ , so $v$ falls to 0 at $y = \frac{h}{2}$ .
Curve from $h$ to the negative $v$ -axis, then from the positive $v$ -axis to $\frac{h}{2}$ : answer (c).

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Motion Under Gravity question from JEE Main 2020 (4 Sep, Shift 1)

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