Problem 44 - Physics for Scientists and Engineers with Modern Physics

Question

\((a)\) A long jumper leaves the ground at \(45^{\circ}\) above the horizontal and lands \(8.0 \mathrm{~m}\) away. What is her "takeoff' speed \(v_{0} ?(b)\) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is \(10.0 \mathrm{~m}\) away horizontally and \(2.5 \mathrm{~m},\) vertically below. If she long jumps from the edge of the left bank at \(45^{\circ}\) with the speed calculated in \((a),\) how long, or short, of the opposite bank will she land (Fig. \(3-43\) )?

Step-by-Step Solution

Answer

(a) Takeoff speed is approximately 8.85 m/s. (b) With this speed, she falls short of the opposite bank.

Step 1: Identify Known Values for Part (a)

In the first part, we need to find the takeoff speed of the long jumper who lands 8.0 m away. The angle of takeoff is given as \(45^{\circ}\). We assume the acceleration due to gravity \(g = 9.8 \, \text{m/s}^2\). The formula for the range \(R\) of a projectile launched at an angle \(\theta\) with initial speed \(v_0\) is: \[ R = \frac{v_0^2 \sin 2\theta}{g} \] We know \(R = 8.0 \, \text{m}\) and \(\theta = 45^{\circ}\). The task is to solve for \(v_0\).

Step 2: Calculate Takeoff Speed

Using the formula from the previous step: \[ R = \frac{v_0^2 \sin 2\theta}{g} = \frac{v_0^2 \sin 90^{\circ}}{g} = \frac{v_0^2}{g} \] Rearranging, we find: \[ v_0^2 = R \cdot g \] Substituting known values, \( R = 8.0 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \), gives: \[ v_0^2 = 8.0 \cdot 9.8 \] Solving, \( v_0^2 = 78.4 \) so \( v_0 = \sqrt{78.4} \approx 8.85 \, \text{m/s} \).

Step 3: Identify Known Values for Part (b)

For the second part, the horizontal distance to the opposite bank is \(10.0 \, \text{m}\) and the vertical drop is \(2.5 \, \text{m}\). Using the speed \(v_0 = 8.85 \, \text{m/s}\) and angle \(\theta = 45^{\circ}\) from the previous part, we need to determine if she can land safely on the other side.

Step 4: Calculate Time of Flight

The time of flight \(t\) in projectile motion is given by the vertical motion equation: \[ y = v_0 \sin \theta \cdot t - \frac{1}{2}gt^2 \] Solving this for \(t\), we set \(y = -2.5 \, \text{m}\) (since she lands below starting level): \[ -2.5 = 8.85 \cdot \sin 45^{\circ} \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2 \] Solving this quadratic equation gives the time of flight \(t\).

Step 5: Calculate Horizontal Distance Travelled

Using the time of flight \(t\) from the previous calculation, calculate the horizontal distance using the formula: \[ x = v_0 \cos \theta \cdot t \] Substituting \(v_0 = 8.85 \, \text{m/s}\) and \(\cos 45^{\circ}\), we find \(x\).

Step 6: Determine Landing Position Relative to Bank

Compare the horizontal distance \(x\) obtained with the 10.0 m horizontal distance to the opposite bank. If \(x > 10.0\), she overshoots, otherwise, she falls short. Calculate by how much or little she reaches or misses the bank.

Key Concepts

Range of a Projectile

The concept of the range of a projectile is crucial in understanding how far an object will travel through the air when launched. The range is determined by both the speed at which the object is launched and the angle of launch. To calculate the range, we use the equation: \[ R = \frac{v_0^2 \sin 2\theta}{g} \] Here, \( R \) represents the range, \( v_0 \) is the initial launch speed, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity, typically \( 9.8 \, \text{m/s}^2 \).

• A launch angle of \( 45^{\circ} \) is ideal for maximizing the range, given constant speed and negligible air resistance.
• Projectiles travel the greatest distance when their velocity is split equally between horizontal and vertical components.
• This equation highlights the direct relationship between the range and the square of the velocity, meaning a small increase in speed can significantly impact distance.

Understanding how different launch angles and speeds alter the range helps in solving complex motion problems and designing better experiments.

Projectile Angle of Launch

The angle at which a projectile is launched plays a vital role in its trajectory and overall performance. When analyzing the angle of launch, we focus on:

• How it affects the initial velocity components: a certain angle divides the initial speed into vertical and horizontal components.
• A \( 45^{\circ} \) angle results in equal distribution between these components, leading to optimal range for level ground launches.
• Angles greater than \( 45^{\circ} \) increase vertical motion but decrease horizontal distance, while angles less than \( 45^{\circ} \) decrease airtime but increase horizontal speed.

This understanding lets us predict not only how far a projectile will travel but also how high it will reach and the time it will remain airborne. By tweaking the angle of launch, you can control the flight of the projectile to achieve specific goals, whether it's to clear an obstacle or reach a precise target.

Time of Flight Calculation

The time that a projectile spends in the air is known as the time of flight. It's determined mainly by vertical motion, given the formula: \[ y = v_0 \sin \theta \cdot t - \frac{1}{2}gt^2 \] Where \( y \) is the vertical displacement, \( t \) is the time of flight, and \( v_0 \) is the initial velocity.

• When launching from and landing on the same level, the time of flight is simpler to compute.
• If the projectile lands below or above the launch point, as in the exercise, the equation needs solving as a quadratic equation.
• Understanding time of flight aids in calculating where the projectile lands and in optimizing where it should land for specific tasks.

This aspect of projectile motion is essential for predicting the landing position and timing of different projectiles launched at varying heights and angles.

Horizontal and Vertical Motion

In projectile motion, the horizontal and vertical components of motion must be considered separately to fully understand the projectile's trajectory.

• Horizontal Motion: This is uniform, meaning there is no acceleration acting in the horizontal direction, assuming air resistance is negligible. It is calculated as \( x = v_0 \cos \theta \cdot t \), where \( x \) is the horizontal distance traveled.
• Vertical Motion: This is affected by gravity, causing the projectile to accelerate downwards. Its motion is described by the initial vertical velocity \( v_0 \sin \theta \) and the gravitational acceleration \( g \).
• The combination of these two motions determines the overall path, or trajectory, of the projectile.

Breaking down projectile motion into horizontal and vertical components allows physicists and engineers to better predict and control the behavior of objects in motion, leading to more precise outcomes in both theoretical calculations and practical applications.