Problem 45 - Physics for Scientists and Engineers with Modern Physics

Question

A high diver leaves the end of a 5.0 -m-high diving board and strikes the water \(1.3 \mathrm{~s}\) later, \(3.0 \mathrm{~m}\) beyond the end of the board. Considering the diver as a particle, determine (a) her initial velocity, \(\overrightarrow{\mathbf{v}}_{0} ;(b)\) the maximum height reached; and \((c)\) the velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{f}}\) with which she enters the water.

Step-by-Step Solution

Answer

Initial velocity: 5.89 m/s; Maximum height: 6.5 m; Final velocity: 7.72 m/s.

Step 1: Identify the Known Values

The diver leaves the diving board at a height of 5.0 m and hits the water after 1.3 seconds, landing 3.0 meters horizontally from the end of the board. The acceleration due to gravity is approximately 9.81 m/s² downward.

Step 2: Decompose the Motion

The motion can be broken into two components: horizontal and vertical. The horizontal motion is uniform because there's no horizontal acceleration, and the vertical motion has constant acceleration due to gravity.

Step 3: Solve for Horizontal Initial Velocity

Using the formula for horizontal displacement \(x = v_{0x} \times t\), we substitute the known values: \(3.0 = v_{0x} \times 1.3\). Solving for \(v_{0x}\), we get \(v_{0x} = \frac{3.0}{1.3} = 2.31\, \mathrm{m/s}\).

Step 4: Solve for Vertical Initial Velocity

Using the vertical motion equation \(y = v_{0y} \times t + \frac{1}{2} g t^2\), and substituting \(y = -5.0\), \(g = 9.81\, \mathrm{m/s^2}\), and \(t = 1.3\), we have \(-5.0 = v_{0y} \times 1.3 - \frac{1}{2} \times 9.81 \times (1.3)^2\). Solving for \(v_{0y}\) gives \(v_{0y} \approx 5.42\, \mathrm{m/s}\).

Step 5: Calculate Initial Velocity Vector

The initial velocity vector is found by combining both components: \(\overrightarrow{v}_{0} = \sqrt{v_{0x}^2 + v_{0y}^2}\). Substituting \(v_{0x} = 2.31\) and \(v_{0y} = 5.42\), \(\overrightarrow{v}_{0} = \sqrt{(2.31)^2 + (5.42)^2} \approx 5.89\, \mathrm{m/s}\).

Step 6: Determine the Maximum Height

The maximum height is found using the vertical motion equation at the peak, where the vertical velocity is 0: \(v_y^2 = v_{0y}^2 - 2g\Delta y\). Solving for \(\Delta y\) gives \(\Delta y = \frac{v_{0y}^2}{2g} = \frac{(5.42)^2}{2 \times 9.81} \approx 1.5 \, \mathrm{m}\). Adding this to the initial height, the maximum height is \(5 + 1.5 = 6.5 \, \mathrm{m}\).

Step 7: Calculate Final Vertical Velocity

Calculating the vertical velocity just before hitting the water using \(v_y = v_{0y} - g \cdot t\), where \(v_y = 5.42 - 9.81 \times 1.3 \approx -7.35 \, \mathrm{m/s}\). The negative sign indicates downward direction.

Step 8: Calculate Final Velocity Vector

The final velocity vector combines the horizontal and final vertical velocities: \(\overrightarrow{v}_{f} = \sqrt{v_{0x}^2 + v_{y}^2}\). Substituting \(v_y = -7.35\) and \(v_{0x} = 2.31\), \(\overrightarrow{v}_{f} = \sqrt{(2.31)^2 + (-7.35)^2} \approx 7.72\, \mathrm{m/s}\).

Key Concepts

Initial Velocity Calculation

In projectile motion, the initial velocity is a combination of horizontal and vertical components. These components are calculated separately because the motion of a projectile is composed of uniform motion in the horizontal direction and uniformly accelerated motion in the vertical direction.

• Horizontal Component: Since there is no acceleration horizontally, the formula for horizontal displacement helps us find the horizontal initial velocity: \(x = v_{0x} \times t\). Given that the diver lands 3.0 meters away from the board in 1.3 seconds, the calculation becomes: \(v_{0x} = \frac{3.0}{1.3} \approx 2.31\, \text{m/s}\).
• Vertical Component: Using the vertical motion equation \(y = v_{0y} \times t + \frac{1}{2} g t^2\), where \(g = 9.81\, \text{m/s}^2\), we account for the vertical drop of 5 meters: \(-5.0 = v_{0y} \times 1.3 - \frac{1}{2} \cdot 9.81 \cdot (1.3)^2\). Solving gives \(v_{0y} \approx 5.42\, \text{m/s}\).

By combining these components, we calculate the overall initial velocity: \(\overrightarrow{v}_{0} = \sqrt{(2.31)^2 + (5.42)^2} \approx 5.89\, \text{m/s}\).

Maximum Height Determination

The maximum height in projectile motion is found when the vertical velocity becomes zero. At this point, the projectile has reached its peak before starting to descend.

To determine this maximum height:

• Use the equation: \(v_y^2 = v_{0y}^2 - 2g\Delta y\), where \(v_y\) is the velocity at the max height, which is 0.
• Rearrange to solve for the change in height \(\Delta y\): \(\Delta y = \frac{v_{0y}^2}{2g}\).
• Substitute the known values: \(\Delta y = \frac{(5.42)^2}{2 \times 9.81} \approx 1.5 \, \text{m}\).

This \(\Delta y\) is added to the original height of the diving board to determine the total maximum height: \(5 + 1.5 = 6.5 \, \text{m}\).

Final Velocity Calculation

The final velocity of a projectile just before impact is a combination of its horizontal and vertical components. These components behave differently under the influence of gravity.

• Horizontal Velocity: This remains constant at \(v_{0x} = 2.31\, \text{m/s}\), as there is no horizontal acceleration.
• Vertical Velocity at Impact: The final vertical velocity \(v_y\) is calculated by considering the initial vertical velocity and the time difference: \(v_y = v_{0y} - g \cdot t\). Plug in the values: \(v_y = 5.42 - 9.81 \times 1.3 \approx -7.35\, \text{m/s}\). The negative implies a downward direction.

Combine these to find the magnitude of the final velocity vector: \(\overrightarrow{v}_{f} = \sqrt{(2.31)^2 + (-7.35)^2} \approx 7.72\, \text{m/s}\).