Problem 21 - Essential Calculus Early Transcendentals

Question

A ball is thrown at an angle of \(45^{\circ}\) to the ground. If the ball lands 90 \(\mathrm{m}\) away, what was the initial speed of the ball?

Step-by-Step Solution

Answer

The initial speed of the ball was approximately 29.7 m/s.

Step 1: Understand the Problem

We need to find the initial speed of a ball that is thrown at a 45-degree angle and lands 90 meters away.

Step 2: Identify Known Variables

The range (horizontal distance) the ball travels is 90 meters, and the angle of launch is 45 degrees. The acceleration due to gravity is approximately \(9.8 \, \text{m/s}^2\).

Step 3: Use Range Formula for Projectile Motion

The formula to calculate the range (\(R\)) of a projectile is given by \[ R = \frac{v_0^2}{g} \sin(2\theta) \]where \(v_0\) is the initial speed, \(g\) is the acceleration due to gravity, and \(\theta\) is the launch angle.

Step 4: Substitute Known Values Into the Formula

We know that \(R = 90 \), \(\theta = 45^{\circ}\), and \(g = 9.8\). Plug these values into the equation:\[ 90 = \frac{v_0^2}{9.8} \sin(90^\circ) \] Since \(\sin(90^\circ) = 1\), the equation simplifies to \[ 90 = \frac{v_0^2}{9.8} \]

Step 5: Solve for Initial Speed \(v_0\)

Rearrange the equation to find \(v_0\):\[ v_0^2 = 90 \times 9.8 \]\[ v_0^2 = 882 \]Take the square root of both sides to solve for \(v_0\):\[ v_0 = \sqrt{882} \approx 29.7 \text{ m/s} \]

Step 6: Verify Calculation

Double-check the input values and calculations to ensure everything is accurate. The initial speed is approximately \(29.7 \, \text{m/s}\).

Key Concepts

Initial Speed Calculation

When trying to find the initial speed of a projectile, we start with the projectile's range, launch angle, and gravity. These are our pieces of the puzzle. The initial speed is the velocity at which the projectile is launched.
To find this, we use the range formula for projectile motion. This formula connects these variables, allowing us to isolate and solve for the initial speed. By observing how far a projectile travels horizontally, we can deduce its initial velocity when combined with knowledge of launch angle and gravitational pull.
The formula to find the range \( R \) is:

• \( R = \frac{v_0^2}{g} \sin(2\theta) \)

Here, \( v_0 \) is the initial speed, \( g \) is the acceleration due to gravity, approximately \(9.8 \ \text{m/s}^2\), and \( \theta \) is the launch angle.
From manipulating this formula, we see that by knowing the range of the projectile and the angle at which it was launched, we can find the initial speed accurately, as shown in the problem solution by isolating \( v_0 \) and solving for it.

Range Formula

The range of a projectile is the total horizontal distance it travels. Understanding this is crucial to solve most related problems.
The projectile's range depends on the following factors:

• Initial speed \( v_0 \)
• Launch angle \( \theta \)
• Acceleration due to gravity \( g \)

The formula for the range \( R \) is given by:

• \( R = \frac{v_0^2}{g} \sin(2\theta) \)

When applied, this equation provides insight into how far a projectile will go given the initial speed and launch angle. Note that it considers the entire motion path, including the upward and downward arcs.
By utilizing this formula, as demonstrated in the problem, we can calculate the necessary initial speed, ensuring the projectile reaches the intended horizontal distance.

Launch Angle Effect

The launch angle significantly influences the projectile's trajectory. It affects both the range and the shape of the path the projectile will take.
To maximize the distance a projectile travels, the optimal launch angle is often \(45^\circ\). This angle provides the perfect balance between height and distance.

• This angle allows for maximum horizontal and vertical velocity components of equal size.

When dealing with different angles, such as those less than \(45^\circ\) or more, it affects how high and how far the projectile travels. The angle changes the \( \sin(2\theta) \) part of the formula, impacting the overall distance.
In our specific exercise, using a \(45^\circ\) launch angle ensures that \( \sin(90^\circ) = 1 \), simplifying calculations. Therefore, understanding the effect of the launch angle is crucial for predicting and maximizing a projectile's range properly.