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Chapter 3: Problem 45

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.

Short Answer

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

Step by step solution

01

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/s虏 downward.
02

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.
03

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 \(v_{0x} = \frac{3.0}{1.3} = 2.31\, \mathrm{m/s}\).
04

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 \(v_{0y} \approx 5.42\, \mathrm{m/s}\).
05

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 \(v_{0y} = 5.42\), \(\overrightarrow{v}_{0} = \sqrt{(2.31)^2 + (5.42)^2} \approx 5.89\, \mathrm{m/s}\).
06

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 \(5 + 1.5 = 6.5 \, \mathrm{m}\).
07

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.
08

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 \(v_{0x} = 2.31\), \(\overrightarrow{v}_{f} = \sqrt{(2.31)^2 + (-7.35)^2} \approx 7.72\, \mathrm{m/s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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: \(\overrightarrow{v}_{f} = \sqrt{(2.31)^2 + (-7.35)^2} \approx 7.72\, \text{m/s}\).

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Most popular questions from this chapter

Huck Finn walks at a speed of \(0.70 \mathrm{~m} / \mathrm{s}\) across his raft (that is, he walks perpendicular to the raft's motion relative to the shore). The raft is traveling down the Mississippi River at a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) relative to the river bank (Fig. \(3-49\) ). What is Huck's velocity (speed and direction) relative to the river bank?

(II) A motorboat whose speed in still water is 3.40 \(\mathrm{m} / \mathrm{s}\) must aim upstream at an angle of \(19.5^{\circ}\) (with respect to a line perpendicular to the shore) in order to travel dircctly across the stream. (a) What is the spced of the current? (b) What is the resultant speed of the boat with respect to the shore?

(II) Suppose the position of an object is given by \(\vec{\mathbf{r}}=\left(3.0 t^{2} \hat{\mathbf{i}}-6.0 t^{3} \hat{\mathbf{j}}\right) \mathrm{m}\) (a) Determine its velocity \(\vec{\mathbf{v}}\) and acceleration \(\vec{a},\) as a function of time. (b) Determine \(\vec{\mathbf{r}}\) and \(\vec{\mathbf{v}}\) at time \(t=2.5 \mathrm{s} .\)

The specd of a boat in still water is \(v\) . The boat is to make a round trip in a river whose current travels at speed u. Derive a formula for the time needed to make a round trip of total distance \(D\) if the boat makes the round trip by moving (a) upstream and back downstream, and \((b)\) directly across the river and back. We must assume \(u

(II) A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with spced 10.0 \(\mathrm{m} / \mathrm{s}\) (Fig. 52\()\) . What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground (a) if the hot-air balloon is rising at 5.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground during this throw, (b) if the hot-air balloon is descending at 5.0 \(\mathrm{m} / \mathrm{s}\) relative to the ground.
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