Q.Galileo, in his book Two new sciences, stated that “for elevations which exceed or fall short of $45^\circ$ by equal amounts, the ranges are equal”. Prove this statement.

We are told a body is projected from the ground at an angle $\theta = \tan^{-1}(\sqrt{7})$. That means $\tan\theta = \sqrt{7}$, so we can find $\sin\theta$ and $\cos\theta$:

The launch speed is $u$. At half the maximum height, the speed is $v = n u$. We need $n$.

Concept & Intuition:

In projectile motion, horizontal velocity is constant (no horizontal force). Vertical motion is governed by gravity, so vertical speed changes with height. Energy conservation is the cleanest way: the loss in kinetic energy equals the gain in gravitational potential energy. At half the maximum height, the potential energy is half of what it would be at the top, so the kinetic energy is reduced by that amount. This directly gives the speed ratio without solving for time.

Step-by-step solution:

1. Maximum height reached: The vertical component of initial velocity is $u_y = u \sin\theta$. At the top, vertical speed is zero. Using $v_y^2 = u_y^2 - 2gH$:

2. Half of maximum height:

   Let $h = \frac{H}{2} = \frac{u^2 \sin^2\theta}{4g}$.

3. Energy conservation between launch and height $h$:

   At launch: total mechanical energy = $\frac{1}{2} m u^2$ (taking ground as zero potential).

   At height $h$: kinetic energy = $\frac{1}{2} m v^2$, potential energy = $m g h$.

   So:

Cancel $m$ and multiply by 2:

4. Substitute $h$: $$$$

Concept and Intuition

The problem gives two scenarios with the same initial speed.

• On level ground, maximum range occurs at a $45^\circ$ launch angle.
• From a tower, the ball is thrown horizontally — so its entire initial velocity is horizontal, and it falls under gravity. The horizontal distance covered is simply the horizontal speed multiplied by the time it takes to hit the ground. We can extract the speed from the first scenario and then apply it to the second.

Step-by-step solution

1. Find the initial speed from the maximum range on level ground. For a projectile launched from ground level, the range is

Maximum range occurs when $\sin 2\theta = 1$ (i.e., $\theta = 45^\circ$), giving

Here $R_{\text{max}} = 32\ \text{m}$ and $g = 10\ \text{m/s}^2$. So

2. Now consider the horizontal throw from the tower. The ball is thrown horizontally with the same speed $u = 8\sqrt{5}\ \text{m/s}$. The tower height is $h = 25\ \text{m}$. The time to fall vertically from rest is found from

Hence

The problem connects two classic projectile scenarios: maximum horizontal range and vertical throw. The crucial insight is that the same initial speed is used in both cases. For a given launch speed $u$, the maximum range on level ground occurs at a $45^\circ$ launch angle, and the maximum height when thrown straight up is simply $u^2/(2g)$. So if we can find $u^2/(2g)$ from the range data, we have the answer directly.

1. Relate maximum range to initial speed. For a projectile launched at angle $\theta$ with speed $u$, the range is

The maximum range occurs when $\sin 2\theta = 1$, i.e., $\theta = 45^\circ$. Thus

We are told $R_{\text{max}} = 80\,\text{m}$, so

2. Find maximum height for vertical throw. When the ball is thrown straight upward with the same speed $u$, the maximum height $H$ is given by

Substitute $u^2 = 80g$:

3. Interpret the result. …

The path followed by a projectile, known as its trajectory, is a parabola. This parabolic path arises from the combination of uniform horizontal motion (constant velocity) and uniformly accelerated vertical motion (due to gravity). To find the equation of this trajectory, we express the horizontal and vertical positions of the projectile as functions of time and then eliminate time.

Here's how we can determine the value of $\gamma$:

1. Identify initial velocity components:

   The initial velocity of the projectile is given as $\vec{u} = (3\hat{i} + 4\hat{j})\, m/s$.

   This means the initial horizontal velocity component is $u_x = 3\, m/s$, and the initial vertical velocity component is $u_y = 4\, m/s$.

   The acceleration due to gravity is $g = 10\, m/s^2$, acting downwards.

2. Formulate equations of motion:

   For horizontal motion, there is no acceleration (assuming air resistance is negligible).

   The horizontal displacement $x$ after time $t$ is given by:

   $x = u_x t$

   $x = 3t \quad \implies \quad t = \frac{x}{3} \quad \text{(Equation 1)}$

   For vertical motion, the acceleration is $-g$ (downwards).

   The vertical displacement $y$ after time $t$ is given by:

   $y = u_y t - \frac{1}{2} g t^2$

   $y = 4t - \frac{1}{2} (10) t^2$

   $y = 4t - 5t^2 \quad \text{(Equation 2)}$

3. Derive the equation of trajectory:

   Substitute Equation 1 into Equation 2 to eliminate $t$:

   $y = 4\left(\frac{x}{3}\right) - 5\left(\frac{x}{3}\right)^2$

   $y = \frac{4x}{3} - 5\left(\frac{x^2}{9}\right)$

   $y = \frac{4x}{3} - \frac{5x^2}{9}$

   The general equation of trajectory for a projectile launched with initial horizontal velocity $u_x$ and initial vertical velocity $u_y$ is:

   $y = \frac{u_y}{u_x} x - \frac{g}{2u_x^2} x^2$ …

The key concept here is the line-of-sight propagation of TV signals. A transmitting tower of height $h$ can only be seen by a receiver up to the point where the Earth’s curvature blocks the line of sight. For a spherical Earth of radius $R$, the maximum distance $d$ (the range) from the tower to the horizon is given by the geometry of a right triangle: the tower height is one leg, the Earth’s radius is another, and the line of sight is the hypotenuse.

From simple geometry, using the approximation $h \ll R$, we get the classic result:

This tells us that $d \propto \sqrt{h}$. So if the height changes, the range changes as the square root of the height ratio.

Now let’s work through the problem step by step.

1. Write the initial range. For height $h$, the range is

2. Write the new range for the increased height. New height $h' = \frac{3}{2}h$. So the new range $d'$ is

3. Express $d'$ in terms of the original $d$. Since $d = \sqrt{2 R h}$, we have

4. Find the change in range. The change is $d' - d$. Substituting: …