Question

From the edge of the rooftop of a building, a boy throws a stone at an angle \(25.0^{\circ}\) above the horizontal. The stone hits the ground 4.20 s later, 105 m away from the base of the building. (Ignore air resistance.) (a) For the stone's path through the air, sketch graphs of \(x, y, v_{x}\) and \(v_{y}\) as functions of time. These need to be only qualitatively correct- you need not put numbers on the axes. (b) Find the initial velocity of the stone. (c) Find the initial height \(h\) from which the stone was thrown. (d) Find the maximum height \(H\) reached by the stone.

Short answer

The initial velocity is approximately 27.57 m/s. The initial height is 21.35 m, and the maximum height reached is 7.54 m.

Step-by-step solution

Understanding the Problem

We have a stone thrown at an angle of \(25.0^{\circ}\) above the horizontal from a building. It lands 105 m away after 4.2 s. We need sketches for the motion and determine the initial velocity, initial height, and maximum height.

Break Down the Components

Express the initial velocity as \(\mathbf{v_{0}} = \begin{pmatrix} v_{0x} \ v_{0y} \end{pmatrix}\). The components are \(v_{0x} = v_{0} \cos(25^{\circ})\) and \(v_{0y} = v_{0} \sin(25^{\circ})\). Use these to analyze vertical and horizontal motions separately.

Horizontal Motion

The horizontal motion is described by \(x = v_{0x} \cdot t\). We know \(x = 105\) m, and \(t = 4.20\) s. Thus, \(v_{0x} = \frac{105}{4.20}\).

Calculate Horizontal Component

Calculate \(v_{0x} = \frac{105}{4.20} \approx 25.0 \, \text{m/s}\). Use \(v_{0x} = v_{0} \cos(25^{\circ})\) to express \(v_{0}\) in terms of \(v_{0x}\).

Vertical Motion Equation

The vertical motion follows \(y = v_{0y} t - \frac{1}{2} gt^2\). We set \(y = -h\) as the stone hits the ground, with \(g = 9.81\, \text{m/s}^{2}\), and \(t = 4.2\) s.

Calculate Initial Vertical Component

Combine the two components: \(v_{0y} = v_{0} \sin(25^{\circ})\). Solve the equation \(-h = v_{0y}(4.2) - \frac{1}{2} \cdot 9.81 \cdot (4.2)^2\). Substitute and solve for \(v_{0y}\).

Calculate Initial Velocity

Use \(v_{0} = \frac{v_{0x}}{\cos(25^{\circ})}\). This gives \(v_{0} = \frac{25.0}{\cos(25^{\circ})} \approx 27.57\, \text{m/s}\).

Find Initial Height

Substitute \(v_{0y} = 27.57 \sin(25^{\circ})\) into the vertical equation and solve for \(h\). This gives \(h = \frac{1}{2} \cdot 9.81 \cdot (4.2)^2 - (27.57 \sin(25^{\circ})) \cdot 4.2 \approx 21.35 \, \text{m}\).

Calculate Maximum Height

The maximum height \(H\) is given by \(v_{0y}^2 = 2gH\). With \(v_{0y} = 27.57 \sin(25^{\circ})\), calculate \(H = \frac{(v_{0y})^2}{2g} \approx 7.54\, \text{m}\).

Sketch Qualitative Graphs

1. \(x(t)\): linearly increasing with time. 2. \(y(t)\): parabolic, starting at \(h\) and descending to zero. 3. \(v_{x}(t)\): constant line as horizontal velocity is constant. 4. \(v_{y}(t)\): linear decrease from \(v_{0y}\) to negative at impact.

Key concepts

Horizontal Motion

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

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.

Initial Velocity Calculation

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.

Maximum Height Calculation

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.