A ball is projected from the ground at an angle of 45 degrees with the horizontal surface. It reache

Q: Wait, why don't we just calculate the exact value of the initial velocity $u$?

You technically could, but it's unnecessary and introduces messy square roots! By keeping the term $\frac{u^2}{g}$ as a single 'package' or variable, we make the algebra much cleaner and reduce the chance of calculation errors. This is a very common and powerful trick in physics problem-solving.

Q: Does the mass of the ball affect the final height?

Not at all! Notice how the mass $m$ appears on both sides of the kinetic energy equation: $\frac{1}{2}mv^2 = \frac{1}{2}(\frac{1}{2}mu^2)$. It completely cancels out. In kinematics, as long as air resistance is ignored, the trajectory is independent of the object's mass.

Q: Why does the angle change from $45^\circ$ to $30^\circ$ after the bounce?

The problem explicitly states the new angle is $30^\circ$. In a real physical scenario, this happens because the collision with the ground is inelastic and involves friction. The vertical component of velocity is reduced due to the loss of kinetic energy, and the horizontal component might also change depending on the surface, resulting in a new launch angle.

Q: Can we use the conservation of mechanical energy for the entire process?

No, we cannot use it for the *entire* process because the collision with the ground is inelastic. Energy is lost to sound, heat, and deformation of the ball. We can only use energy conservation *during* the flight (from launch to just before hitting the ground), but we must use the given condition ($K_f = \frac{1}{2}K_i$) to bridge the gap across the collision.

Q: What would the second height be if the collision was perfectly elastic?

If the collision were perfectly elastic, no kinetic energy would be lost, meaning the speed $v$ would equal $u$. If the angle somehow still changed to $30^\circ$, the new height would be $h = \frac{u^2 \sin^2 30^\circ}{2g} = \frac{u^2 (1/4)}{2g} = \frac{u^2}{8g}$. Since $\frac{u^2}{4g} = 120$, $\frac{u^2}{8g}$ would be $60\text{ m}$.

Question

A ball is projected from the ground at an angle of $45^\circ$ with the horizontal surface. It reaches a maximum height of $120	ext{ m}$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^\circ$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is .......... .

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Flights

Imagine a ball launched into the air. It traces a beautiful parabolic path, reaches a peak of $120	ext{ m}$, and crashes back to the ground. But the story doesn't end there. It bounces back, but with less energy, and at a shallower angle. Our goal is to find the peak of this second, smaller bounce.

This problem is a classic blend of kinematics and work-energy principles. It tests not just your ability to plug numbers into formulas, but your ability to link two distinct physical states through a collision.

### Phase 1: The Initial Ascent

The formula for the maximum height $H$ of a projectile is one of the most elegant results in kinematics:

$$H = \frac{u^2 \sin^2 	heta}{2g}$$

Let's apply this to our first flight. We know the maximum height is $120	ext{ m}$ and the launch angle is $45^\circ$. Substituting these values into our master equation, we get:

$$120 = \frac{u^2 \sin^2 45^\circ}{2g}$$

We know that $\sin 45^\circ = \frac{1}{\sqrt{2}}$, which means $\sin^2 45^\circ = \frac{1}{2}$. Plugging this in:

$$120 = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g}$$

**Here is a crucial problem-solving secret:** Do not try to solve for $u$ or $g$ individually! We don't need them. Instead, treat the entire expression $\frac{u^2}{4g}$ as a single "package" or variable. We know this package is exactly equal to $120$. We will save this for later.

### The Collision: A Toll is Paid

When the ball hits the ground, it undergoes an inelastic collision. The problem states that it loses half of its kinetic energy.

Kinetic energy is given by $K = \frac{1}{2}mv^2$. If the final kinetic energy $K_f$ is half of the initial kinetic energy $K_i$, we can write:

$$K_f = \frac{1}{2} K_i$$

$$\frac{1}{2} m v^2 = \frac{1}{2} \left( \frac{1}{2} m u^2 ight)$$

Notice how the mass $m$ and the leading $\frac{1}{2}$ cancel out perfectly from both sides. This leaves us with a direct relationship between the squares of the velocities:

$$v^2 = \frac{u^2}{2}$$

This tells us that the square of the launch speed for the second flight is exactly half the square of the launch speed for the first flight.

### Phase 2: The Rebound

Now, let's analyze the second flight. The ball is launched with a new velocity $v$ at a new angle of $30^\circ$. We want to find the new maximum height, $h$. We use our trusty height formula again:

$$h = \frac{v^2 \sin^2 30^\circ}{2g}$$

Let's substitute what we know. We found that $v^2 = \frac{u^2}{2}$, and we know that $\sin 30^\circ = \frac{1}{2}$, which means $\sin^2 30^\circ = \frac{1}{4}$.

$$h = \frac{\left(\frac{u^2}{2}ight) \left(\frac{1}{4}ight)}{2g}$$

Multiplying the terms in the numerator gives us $\frac{u^2}{8}$. Dividing that by $2g$ yields:

$$h = \frac{u^2}{16g}$$

### The Grand Finale: Connecting the Dots

We have an expression for $h$, but it's in terms of $u$ and $g$. This is where our saved "package" comes to the rescue! We know from Phase 1 that $\frac{u^2}{4g} = 120$.

Let's rewrite our expression for $h$ to expose this package:

$$h = \frac{u^2}{16g} = \frac{1}{4} \left( \frac{u^2}{4g} ight)$$

Now, we simply substitute $120$ for the package:

$$h = \frac{1}{4} (120) = 30	ext{ m}$$

The ball reaches a maximum height of $30	ext{ m}$ on its second bounce.

### An Elegant Alternative: The Vertical Velocity Perspective

There is an even faster, more intuitive way to think about this problem. The maximum height of any projectile depends *only* on its initial vertical velocity, $u_y$. The formula can be written as:

$$H = \frac{u_y^2}{2g}$$

Let's look at the vertical velocity for both flights.

For the first flight, the vertical velocity is:
$$u_y = u \sin 45^\circ = \frac{u}{\sqrt{2}}$$

For the second flight, the new launch speed is $v = \frac{u}{\sqrt{2}}$, and the new angle is $30^\circ$. So, the new vertical velocity is:
$$v_y = v \sin 30^\circ = \left(\frac{u}{\sqrt{2}}ight) \left(\frac{1}{2}ight) = \frac{u}{2\sqrt{2}}$$

Notice the relationship between $u_y$ and $v_y$? The new vertical velocity $v_y$ is exactly **half** of the original vertical velocity $u_y$!

$$v_y = \frac{1}{2} u_y$$

Since the maximum height is proportional to the **square** of the vertical velocity, halving the vertical velocity means the maximum height will be reduced to **one-fourth** of its original value.

$$h = \frac{1}{4} H = \frac{1}{4} (120) = 30	ext{ m}$$

This is the true power of physics intuition. By understanding the core dependencies of the equations, we can bypass tedious algebra and arrive at the answer with pure logic!

Enter Numerical Value:

Visualized Solution

The Two Trajectories

First Maximum Height

Simplifying the First Equation

The Energy Loss

Velocity After Bounce

Second Maximum Height

Calculating the Final Height

The Final Substitution

The Sigma Insight: Projectile Motion

The Setup

Phase 1

The Collision

Phase 2

The Grand Finale

An Elegant Alternative

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