A ball is dropped from a height of $90\ \text{m}$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between $t = 0$ to $12\ \text{s}$.

> [!TLDR] > > The ball undergoes a series of free falls and inelastic bounces, losing 1/10th of its speed at each collision. This results in a speed-time graph composed of linear segments with slopes $\pm g$, where speed instantaneously drops at each collision, forming a series of decreasing V-shapes, ending at $t=12\ \text{s}$ with a speed of approximately $33.16\ \text{m/s}$. The problem asks us to plot the speed-time graph of a ball dropped from a height, where it loses a fraction of its speed at each collision with the floor. This scenario involves understanding kinematics under constant gravitational acceleration and the effect of inelastic collisions. ### Concept and Intuition The motion of the ball is a sequence of vertical free falls and ascents. 1. **Free Fall (Downward Motion):** When the ball is falling, its speed increases linearly due to gravity. If it starts from rest, its speed $v$ at time $t$ is given by $v = gt$. The displacement $s$ is $s = \frac{1}{2}gt^2$. 2. **Vertical Ascent (Upward Motion):** After bouncing, the ball moves upwards. Its speed decreases linearly due to gravity, eventually becoming zero at its maximum height. If it starts with an upward speed $v_0$, its speed $v$ at time $t$ is $v = v_0 - gt$. The time to reach maximum height is $t_{max} = v_0/g$, and the maximum height reached is $H_{max} = v_0^2/(2g)$. 3. **Inelastic Collision with Floor:** This is the critical aspect. The problem states that the ball loses one-tenth of its speed at each collision. This means if the speed just before collision is $v_{\text{before}}$, the speed just after collision $v_{\text{after}}$ will be $v_{\text{after}} = v_{\text{before}} - \frac{1}{10}v_{\text{before}} = \frac{9}{10}v_{\text{before}}$. This loss of speed signifies a loss of kinetic energy, making the collision inelastic. 4. **Consequences of Speed Loss:** Because the ball rebounds with less speed, it will not reach its original height. The maximum height achieved after each subseq

A ball is dropped from a height of 90m on a floor. At each collision with the floor, the ball loses one tenth

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Concept understanding — Motion Under Gravity

Motion Under Gravity – The Intuition First

Drop a ball. It falls. Why? Because the Earth pulls it. But the key insight is not just that it falls — it's how it falls. If you drop a stone from a cliff, does it move at a constant speed? No. It gets faster as it goes down. That's the heart of motion under gravity: the pull is steady, so the acceleration is steady.

Now imagine throwing a ball straight up. It rises, slows down, stops for an instant at the top, then falls back down. The same pull that made it slow down on the way up makes it speed up on the way down. The acceleration never changes direction — it always points downward, toward the centre of the Earth.

That constant downward acceleration is called $g$ . Near the Earth's surface, its magnitude is about $9.8\ \text{m/s}^2$ (often taken as $10\ \text{m/s}^2$ in exams for quick calculation). The direction is always downward.

Important

In one-dimensional motion under gravity, the acceleration is constant and equal to $g$ downward. This is true only when air resistance is negligible and the height is small compared to the Earth's radius.

The Precise Statement

Motion under gravity (in one dimension) means an object moves vertically under the sole influence of the Earth's gravitational pull. The acceleration is:

a = -g

where the negative sign indicates downward direction (if we take upward as positive). The value of $g$ is $9.8\ \text{m/s}^2$ (or $10\ \text{m/s}^2$ in many problems).

Because acceleration is constant, all three equations of uniformly accelerated motion apply directly:

v = u + at

s = ut + \frac12 at^2

v^2 = u^2 + 2as

But here $a$ is replaced by $-g$ (if upward is positive) or $+g$ (if downward is positive). The choice of sign convention is yours — just be consistent.

v = u - gt

Key Features You Must Know

1. Free Fall (dropped from rest)

If you simply let go of an object ( $u = 0$ ), it falls with increasing speed. After time $t$ , its velocity is $v = gt$ (downward). The distance fallen is $s = \frac12 gt^2$ .

2. Projected Upward

If you throw a ball upward with speed $u$ , it rises until its velocity becomes zero at the highest point. That takes time $t = u/g$ . The maximum height reached is $h = u^2/(2g)$ .

3. Symmetry of the Motion

The time to go up equals the time to come back down (to the same height). The speed at which it returns to the launch point equals the initial speed $u$ (but downward). This symmetry is a direct consequence of constant acceleration.

Watch out

At the highest point, velocity is zero but acceleration is still $g$ downward. Many beginners think acceleration becomes zero at the top — it does not. The object is momentarily at rest, but gravity is still pulling it.

A Simple Example …

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The ball undergoes a series of free falls and inelastic bounces, losing 1/10th of its speed at each collision. This results in a speed-time graph composed of linear segments with slopes $\pm g$ , where speed instantaneously drops at each collision, forming a series of decreasing V-shapes, ending at $t=12\ \text{s}$ with a speed of approximately $33.16\ \text{m/s}$ .

The problem asks us to plot the speed-time graph of a ball dropped from a height, where it loses a fraction of its speed at each collision with the floor. This scenario involves understanding kinematics under constant gravitational acceleration and the effect of inelastic collisions.

Concept and Intuition

The motion of the ball is a sequence of vertical free falls and ascents.

Free Fall (Downward Motion): When the ball is falling, its speed increases linearly due to gravity. If it starts from rest, its speed $v$ at time $t$ is given by $v = gt$ . The displacement $s$ is $s = \frac{1}{2}gt^2$ .
Vertical Ascent (Upward Motion): After bouncing, the ball moves upwards. Its speed decreases linearly due to gravity, eventually becoming zero at its maximum height. If it starts with an upward speed $v_0$ , its speed $v$ at time $t$ is $v = v_0 - gt$ . The time to reach maximum height is $t_{max} = v_0/g$ , and the maximum height reached is $H_{max} = v_0^2/(2g)$ .
Inelastic Collision with Floor: This is the critical aspect. The problem states that the ball loses one-tenth of its speed at each collision. This means if the speed just before collision is $v_{\text{before}}$ , the speed just after collision $v_{\text{after}}$ will be $v_{\text{after}} = v_{\text{before}} - \frac{1}{10}v_{\text{before}} = \frac{9}{10}v_{\text{before}}$ . This loss of speed signifies a loss of kinetic energy, making the collision inelastic.
Consequences of Speed Loss: Because the ball rebounds with less speed, it will not reach its original height. The maximum height achieved after each subsequent bounce will be lower. Consequently, the time taken for each subsequent ascent and descent cycle will also be shorter.
Speed-Time Graph Characteristics:
- Since speed is the magnitude of velocity, the graph will always be above or on the time axis (speed is non-negative).
- During free fall, speed increases linearly with time, so the graph segment will have a positive slope ( $+g$ ).
- During vertical ascent, speed decreases linearly with time, so the graph segment will have a negative slope ( $-g$ ).
- At the moment of collision, there is an instantaneous drop in speed, represented by a vertical line segment on the graph.
- The overall graph will be a series of "V" shapes, with each subsequent "V" being shorter and narrower due to the energy loss.

We will use the standard acceleration due to gravity $g = 10\ \text{m/s}^2$ for calculations, as is common in many Indian competitive exams unless specified otherwise. We will also use $\sqrt{2} \approx 1.414$ for numerical approximations.

Step-by-Step Solution

We need to track the ball's speed and time through each phase of its motion until $t = 12\ \text{s}$ .

1. First Fall (from $t=0$ until the first collision)

The ball is dropped from a height of $H_0 = 90\ \text{m}$ .

Time to hit the floor ( $t_1$ ): Using the equation $H = ut + \frac{1}{2}gt^2$ , with $u=0$ :

90 = 0 \cdot t_1 + \frac{1}{2}(10)t_1^2

90 = 5t_1^2 \implies t_1^2 = 18 \implies t_1 = \sqrt{18} = 3\sqrt{2}\ \text{s}

Numerically, $t_1 \approx 3 \times 1.414 = 4.242\ \text{s}$.

Speed just before the first collision ( $v_1$ ): Using $v = u + gt$ :

v_1 = 0 + (10)(3\sqrt{2}) = 30\sqrt{2}\ \text{m/s}

Numerically, $v_1 \approx 30 \times 1.414 = 42.42\ \text{m/s}$.

This segment of the graph goes from $(0, 0)$ to $(4.242\ \text{s}, 42.42\ \text{m/s})$ .

2. First Rebound and Ascent (after the first collision)

Speed just after the first collision ( $v_1'$ ): The ball loses one-tenth of its speed.

v_1' = v_1 - \frac{1}{10}v_1 = \frac{9}{10}v_1 = \frac{9}{10}(30\sqrt{2}) = 27\sqrt{2}\ \text{m/s}

Numerically, $v_1' \approx 27 \times 1.414 = 38.178\ \text{m/s}$.
At $t_1 \approx 4.242\ \text{s}$, the speed instantaneously drops from $42.42\ \text{m/s}$ to $38.178\ \text{m/s}$.

Time to reach maximum height ( $t_{\text{asc1}}$ ): The ball moves upwards with initial speed $v_1'$ . At maximum height, its speed is $0$ . Using $v = u - gt$ :

0 = v_1' - gt_{\text{asc1}} \implies t_{\text{asc1}} = \frac{v_1'}{g} = \frac{27\sqrt{2}}{10} = 2.7\sqrt{2}\ \text{s}

Numerically, $t_{\text{asc1}} \approx 2.7 \times 1.414 = 3.8178\ \text{s}$.

Total time at the peak of the first bounce ( $T_1$ ):

T_1 = t_1 + t_{\text{asc1}} = 3\sqrt{2} + 2.7\sqrt{2} = 5.7\sqrt{2}\ \text{s}

Numerically, $T_1 \approx 5.7 \times 1.414 = 8.0598\ \text{s}$.

This segment of the graph goes from $(4.242\ \text{s}, 38.178\ \text{m/s})$ to $(8.0598\ \text{s}, 0\ \text{m/s})$ .

3. Second Fall (from $T_1$ until the second collision)

The ball falls from rest at height $H_1$ (reached after the first bounce). The time taken to fall from $H_1$ is equal to the time taken to ascend to $H_1$ .

Time for the second fall ( $t_{\text{fall2}}$ ):

t_{\text{fall2}} = t_{\text{asc1}} = 2.7\sqrt{2}\ \text{s}

Numerically, $t_{\text{fall2}} \approx 3.8178\ \text{s}$.

Speed just before the second collision ( $v_2$ ): This speed will be equal to the speed with which it started its first ascent ( $v_1'$ ).

v_2 = gt_{\text{fall2}} = 10(2.7\sqrt{2}) = 27\sqrt{2}\ \text{m/s}

Numerically, $v_2 \approx 38.178\ \text{m/s}$.

Total time at the second collision ( $T_2$ ):

T_2 = T_1 + t_{\text{fall2}} = 5.7\sqrt{2} + 2.7\sqrt{2} = 8.4\sqrt{2}\ \text{s}

Numerically, $T_2 \approx 8.4 \times 1.41421 = 11.88\ \text{s}$.

This segment of the graph goes from $(8.06\ \text{s}, 0\ \text{m/s})$ to $(11.88\ \text{s}, 38.18\ \text{m/s})$ .

4. Second Rebound and Ascent (partial, up to $t=12\ \text{s}$ )

The second collision occurs at $T_2 \approx 11.88\ \text{s}$ . We need to plot the graph up to $t=12\ \text{s}$ .

Speed just after the second collision ( $v_2'$ ):

v_2' = \frac{9}{10}v_2 = \frac{9}{10}(27\sqrt{2}) = 24.3\sqrt{2}\ \text{m/s}

Numerically, $v_2' \approx 24.3 \times 1.41421 = 34.37\ \text{m/s}$.
At $T_2 \approx 11.88\ \text{s}$, the speed instantaneously drops from $38.18\ \text{m/s}$ to $34.37\ \text{m/s}$.

Time remaining for ascent:

\Delta t = 12\ \text{s} - T_2 = 12 - 8.4\sqrt{2}\ \text{s}

Numerically, $\Delta t \approx 12 - 11.8794 = 0.1206\ \text{s}$.

Speed at $t=12\ \text{s}$ ( $v(12)$ ): The ball is ascending with initial speed $v_2'$ . …

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Concept: A Closed-Form (Geometric Series) Formula for Repeated Inelastic Bounces

Method: General Recursive/Closed-Form Formula, Evaluated at $n=1,2$ — Not Step-by-Step Arithmetic

Each bounce multiplies the ball's speed by a fixed ratio $r = 9/10$ (loses one-tenth of its speed). Because the time between successive bounces also scales by this same ratio, the cumulative time up to the $n$ -th bounce is a geometric series with a known closed form — letting you jump straight to "which bounce interval contains $t=12\ \text{s}$ " instead of computing each bounce's numbers in sequence.

Steps

Time and speed of the very first impact. Dropped from $H=90\ \text{m}$ , $g=10\ \text{m s}^{-2}$ :

t_1 = \sqrt{\frac{2H}{g}} = \sqrt{18} = 3\sqrt{2}\ \text{s} \;(\approx 4.243\ \text{s}), \qquad v_1 = gt_1 = 30\sqrt{2}\ \text{m s}^{-1}

Set up the recursion. After the $k$ -th bounce, rebound speed is $r^k v_1$ (where $v_1$ is the impact speed at the first bounce, and $r=0.9$ ). The time for the "up-then-down" cycle following the $k$ -th bounce is $2\times(\text{rebound speed})/g = \dfrac{2r^k v_1}{g} = 2 r^k t_1$ .
Sum the geometric series to get the time of the $n$ -th bounce, $T_n$ :

T_n = t_1 + \sum_{k=1}^{n-1} 2r^k t_1 = t_1\left[1 + 2r\cdot\frac{1-r^{n-1}}{1-r}\right]

This single formula gives the time of any bounce directly — no need to add up each preceding cycle one at a time.

Evaluate for $n=2$ (the second bounce), with $r=0.9$ , $t_1\approx4.2426\ \text{s}$ :

T_2 = t_1[1+2r] = 4.2426\times 2.8 \approx 11.88\ \text{s}

Evaluate for $n=3$ to bracket $t=12\ \text{s}$ :

T_3 = t_1[1+2r(1+r)] = t_1\times 4.42 \approx 18.75\ \text{s}

Since $T_2\approx11.88 < 12 < T_3\approx18.75$ , the instant $t=12\ \text{s}$ falls in the rising (post-2nd-bounce) phase. …

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Q.A ball is dropped from a height of 90 m90\ \text{m}90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t=0t = 0t=0 to 12 s12\ \text{s}12 s.

Concept and Intuition

Step-by-Step Solution

1. First Fall (from t=0t=0t=0 until the first collision)

2. First Rebound and Ascent (after the first collision)

3. Second Fall (from T1T_1T1​ until the second collision)

4. Second Rebound and Ascent (partial, up to t=12 st=12\ \text{s}t=12 s)

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Q.A ball is dropped from a height of $90\ \text{m}$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between $t = 0$ to $12\ \text{s}$ .

1. First Fall (from $t=0$ until the first collision)

3. Second Fall (from $T_1$ until the second collision)

4. Second Rebound and Ascent (partial, up to $t=12\ \text{s}$ )

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