91Ó°ÊÓ

In projectile motion, horizontal motion is a critical component. It takes place at a constant velocity since there is no acceleration acting in the horizontal direction (ignoring air resistance, as in our problem). This is why the path is linear.
When a stone is thrown, its horizontal motion can be mathematically represented by the equation:

• \[ x = v_{0x} \cdot t \]

Here, \(x\) is the distance traveled horizontally, \(v_{0x}\) is the initial horizontal velocity, and \(t\) is time.
This formula shows that the horizontal distance (\(x\) = 105 m) directly depends on the time (4.2 s) it is in motion. From the solution, we know \(v_{0x} \approx 25.0\, \text{m/s}\). This tells us that the horizontal speed does not change during the stone's flight.

Vertical motion involves acceleration due to gravity. The vertical component is more complex due to this downward force. It is often described with the equation:

• \[ y = v_{0y} t - \frac{1}{2} g t^2 \]

Here, \(y\) represents the vertical displacement (from initial height \(h\) to ground), \(v_{0y}\) is the initial vertical velocity, \(g\) is the acceleration due to gravity (9.81 m/s²), and \(t\) is time. In this context, \(y = -h\) when the stone hits the ground.
The calculation is set up to find \(v_{0y}\) and then used to determine initial height \(h\). The concept emphasizes how the vertical motion is influenced by the initial velocity and gravitational pull, showing a descending parabolic path.

Calculating the initial velocity is like piecing a puzzle. It requires breaking down the launch velocity into horizontal and vertical components because the stone was thrown at an angle \(25.0^{\circ}\). The total initial velocity \(v_0\) is given by:

• \[ v_0 = \sqrt{v_{0x}^2 + v_{0y}^2} \]
• \[ v_{0x} = v_0 \cos(\theta) \] and \[ v_{0y} = v_0 \sin(\theta) \]

Using \(v_{0x} \approx 25.0\, \text{m/s}\), we set up for \(v_0\) calculation:

• \[ v_0 = \frac{v_{0x}}{\cos(25^\circ)} \approx 27.57\, \text{m/s} \]

This calculation means finding \(v_0\), which summarizes both horizontal and vertical components to depict the stone's initial thrust into the sky.

Determining the stone's maximum height involves using the initial vertical velocity. Maximum height is where the stone briefly stops moving upward before descending. This occurs when the vertical velocity is zero. The formula is:

• \[ H = \frac{(v_{0y})^2}{2g} \]

Plugging in \(v_{0y} = 27.57 \sin(25^\circ)\), we found the maximum height \(H\) to be approximately 7.54 m. This represents the total vertical distance it traveled from the rooftop before changing direction and demonstrates gravity's role in halting its ascent.