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

Q: Why don't we need to know the exact value of gravity (g) to solve this?

Because we are essentially comparing two states of the same system. By expressing the final height as a ratio of the initial height, the constant 'g' cancels out entirely. This is a powerful technique in physics to simplify calculations.

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

No, it doesn't. While kinetic energy depends on mass, the problem states that a fraction (half) of the energy is lost. When you set up the energy equation, the mass 'm' appears on both sides and cancels out, leaving only a relationship between the velocities.

Q: Why did the angle change from 45 degrees to 30 degrees?

During an inelastic collision with the ground, the vertical and horizontal components of velocity can be affected differently (e.g., due to friction or the coefficient of restitution). This unequal change alters the overall direction of the velocity vector. However, the problem simplifies this by directly giving us the final angle.

Q: Could I have solved this by calculating the exact initial velocity 'u' first?

Yes, you could assume $g = 9.8$ or $10\text{ m/s}^2$, calculate $u$, then calculate $v$, and finally calculate $H_2$. However, doing so introduces messy square roots and increases the chance of calculation errors. Using algebraic ratios is much cleaner and faster.

Q: What if the ball hadn't lost any kinetic energy?

If the collision were perfectly elastic and frictionless, it wouldn't lose kinetic energy, and the angle of reflection would equal the angle of incidence ($45^\circ$). In that case, the ball would bounce back to the exact same maximum height of $120\text{ m}$.

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 Physics of a Bouncing Projectile

Imagine standing on a vast, flat field and launching a ball into the air. It traces a beautiful parabolic arc, reaching a peak before gravity pulls it back down. But the story doesn't end when it hits the ground. The ball bounces, but it doesn't bounce perfectly. It loses energy, and its trajectory changes. This problem is a classic exploration of how kinematics and energy conservation intertwine.

### Analyzing the First Flight

Let's start by looking at the initial launch. The ball is projected at an angle of $45^\circ$ and reaches a maximum height, which we'll call $H_1$, of $120	ext{ m}$.

In projectile motion, the maximum height is determined entirely by the vertical component of the initial velocity. The formula for maximum height is:

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

Here, $u$ is the initial velocity, $	heta$ is the angle of projection, and $g$ is the acceleration due to gravity. Let's plug in what we know for the first flight:

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

We know that $\sin 45^\circ = \frac{1}{\sqrt{2}}$, so squaring it gives us $\frac{1}{2}$. Substituting this back into our equation:

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

This is a crucial checkpoint. We have found a relationship between the initial velocity squared and gravity: $\frac{u^2}{4g} = 120$. We don't need to find the exact value of $u$ or $g$; this ratio is all we need to unlock the rest of the problem.

### The Collision and Energy Loss

When the ball hits the ground, it undergoes an inelastic collision. The problem states that it loses exactly 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:

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

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

Since the mass $m$ of the ball remains constant, it cancels out from both sides. This leaves us with a direct relationship between the velocities:

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

This tells us that the square of the new velocity is exactly half the square of the old velocity.

### The Second Flight

Now, the ball rebounds with this new velocity $v$. The problem also tells us that the new angle of projection is $30^\circ$. We need to find the new maximum height, $H_2$.

We use the same maximum height formula, but with our new parameters:

$$H_2 = \frac{v^2 \sin^2 30^\circ}{2g}$$

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

$$H_2 = \frac{\left(\frac{u^2}{2}ight) \left(\frac{1}{4}ight)}{2g}$$

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

$$H_2 = \frac{u^2}{16g}$$

### The Power of Ratios

Here is where the elegance of physics shines. We have an expression for $H_2$ in terms of $u^2$ and $g$. But remember our checkpoint from the first flight? We established that $\frac{u^2}{4g} = 120$.

Let's rewrite our expression for $H_2$ to reveal this hidden ratio:

$$H_2 = \frac{1}{4} \left( \frac{u^2}{4g} ight)$$

By factoring out the $\frac{1}{4}$, we perfectly isolate the term $\frac{u^2}{4g}$, which is exactly $H_1$. This means the new height is simply one-quarter of the original height!

$$H_2 = \frac{1}{4} H_1$$

$$H_2 = \frac{120}{4} = 30	ext{ m}$$

And there we have it. By tracking the energy loss and understanding how the components of the height formula interact, we arrived at the final answer of $30	ext{ m}$ without ever needing a calculator to find the exact initial speed.

Enter Numerical Value:

Visualized Solution

$H_{1} = \frac{u ^{2} s i n ^{2} θ _{1}}{2 g}$

$\frac{u ^{2}}{4 g} = 120$

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

$H_{2} = \frac{v ^{2} s i n ^{2} θ _{2}}{2 g}$

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

$H_{2} = \frac{H _{1}}{4}$

$H_{2} = 30 m$

The Sigma Insight: Projectile Motion

The Physics of a Bouncing Projectile

Analyzing the First Flight

The Collision and Energy Loss

The Second Flight

The Power of Ratios

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H1​=2gu2sin2θ1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4gu2​=120

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2=2u2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​=2gv2sin2θ2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​=16gu2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​=4H1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​=30 m