A ball is thrown from ground at an angle theta with horizontal and with an initial speed u0. For the

Q: Why is the average velocity for the first bounce just $u_0 \cos\theta$?

Average velocity is total displacement over total time. Since the ball starts and ends on the ground, its vertical displacement is zero. The only displacement is horizontal, which occurs at a constant velocity of $u_0 \cos\theta$.

Q: Why does the displacement scale by $1/\alpha^2$ instead of $1/\alpha$?

Displacement (range) depends on both the horizontal velocity and the time of flight. Since both scale down by $1/\alpha$, their product scales down by $1/\alpha \times 1/\alpha = 1/\alpha^2$.

Q: How do we know there are infinite bounces?

The problem states the motion 'continues for a long time'. In physics, when a geometric progression of bounces occurs, it mathematically converges to a finite time and distance, allowing us to model it as an infinite series.

Q: What if the angle $\theta$ changed after the bounce?

If the angle changed, the horizontal and vertical components would scale differently, and we couldn't use a single scaling factor $\alpha$ for both time and horizontal velocity. The constant angle is a crucial simplifying constraint.

Q: Is $\alpha$ related to the coefficient of restitution?

Yes! The coefficient of restitution $e$ is the ratio of relative velocity of separation to approach. Here, the vertical speed becomes $u_y/\alpha$, so $e = 1/\alpha$. The horizontal speed also changing implies friction or a specific type of inelastic collision, but mathematically, $e = 1/\alpha$.

Question

A ball is thrown from ground at an angle $	heta$ with horizontal and with an initial speed $u_0$. For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is $V_1$. After hitting the ground, ball rebounds at the same angle $	heta$ but with a reduced speed of $u_0/\alpha$. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is $0.8 V_1$, the value of $\alpha$ is______.

Sigma Percentile · Accepted Answer

The beauty of physics often lies in how it seamlessly intertwines with pure mathematics. This problem is a classic example of that harmony, blending the kinematics of projectile motion with the elegance of infinite geometric progressions. Let's embark on this journey step by step.

### Analyzing the First Bounce

Imagine the ball being launched for the very first time. It flies through the air and lands back on the ground. Because it starts and ends at the same horizontal level, its net vertical displacement is exactly zero.

When we talk about the average velocity $V_1$ for this first trip, we are simply looking at the total displacement divided by the total time. Since the only displacement is horizontal, and the horizontal velocity remains constant throughout the flight, the average velocity is just the horizontal component of the initial velocity:

$$V_1 = u_0 \cos	heta$$

The time of flight for this first arc is given by the standard formula $t_1 = \frac{2u_0 \sin	heta}{g}$, and the horizontal range is $d_1 = V_1 t_1$.

### The Scaling Factor of Subsequent Bounces

Now, the ball hits the ground and rebounds. The problem states it rebounds at the same angle $	heta$, but its speed is reduced by a factor of $\alpha$. This means the new initial speed is $u_0/\alpha$.

Because both the horizontal and vertical components of the velocity scale down by this exact factor $\alpha$, everything else scales proportionally. The new average velocity for the second bounce becomes $V_2 = V_1/\alpha$, and the new time of flight becomes $t_2 = t_1/\alpha$.

But what about the horizontal displacement (the range) for the second bounce? Distance is velocity multiplied by time. Since both the velocity and the time have scaled down by $\alpha$, their product scales down by $\alpha^2$:

$$d_2 = V_2 t_2 = \left(\frac{V_1}{\alpha}ight) \left(\frac{t_1}{\alpha}ight) = \frac{d_1}{\alpha^2}$$

This is a crucial realization! The time scales by $1/\alpha$, but the distance scales by $1/\alpha^2$.

### Summing the Infinite Series

The motion continues for a "long time," which in physics parlance means we can model it as an infinite series. To find the overall average velocity, we need the total time $T$ and the total displacement $D$.

The total time is an infinite geometric progression:

$$T = t_1 + \frac{t_1}{\alpha} + \frac{t_1}{\alpha^2} + \dots = \frac{t_1}{1 - \frac{1}{\alpha}}$$

Similarly, the total horizontal displacement is another infinite geometric progression, but with a different common ratio:

$$D = d_1 + \frac{d_1}{\alpha^2} + \frac{d_1}{\alpha^4} + \dots = \frac{d_1}{1 - \frac{1}{\alpha^2}}$$

### The Grand Synthesis

Now for the master stroke. The overall average velocity $V_{avg}$ is the total displacement divided by the total time:

$$V_{avg} = \frac{D}{T} = \frac{\frac{d_1}{1 - 1/\alpha^2}}{\frac{t_1}{1 - 1/\alpha}}$$

We know that $d_1/t_1$ is simply $V_1$. The remaining algebraic terms simplify beautifully because $(1 - 1/\alpha^2)$ is a difference of squares and factors into $(1 - 1/\alpha)(1 + 1/\alpha)$. The $(1 - 1/\alpha)$ terms cancel out perfectly:

$$V_{avg} = V_1 \frac{1}{1 + 1/\alpha} = V_1 \frac{\alpha}{\alpha + 1}$$

We are given that this overall average velocity is $0.8 V_1$, which is $4/5$ of $V_1$. Equating our expression to this value:

$$\frac{\alpha}{\alpha + 1} = \frac{4}{5}$$

Cross-multiplying yields $5\alpha = 4\alpha + 4$, which immediately gives us our final answer:

**$\alpha = 4$**

This problem elegantly masks the concept of the coefficient of restitution $e$. Physically, the speed reducing by a factor of $\alpha$ implies $e = 1/\alpha$. If you ever encounter a similar problem given in terms of $e$, you now know the overall average velocity is simply $V_1 / (1+e)$!

Enter Numerical Value:

Visualized Solution

$First Bounce Analysis$

$Subsequent Bounces$

$Displacement of Bounces$

$Total Time of Flight$

$Total Displacement$

$Overall Average Velocity$

$Solving for α$

$The Way Forward$

The Sigma Insight: Projectile Motion

Analyzing the First Bounce

The Scaling Factor of Subsequent Bounces

Summing the Infinite Series

The Grand Synthesis

Similar Questions

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Bounce Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Subsequent Bounces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Displacement of Bounces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Time of Flight

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Displacement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Overall Average Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward