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Chapter 3: Problem 181

A projectile is fired from ground level at an angle of \(8^{\circ}\) with the horizontal. The projectile is to have a range of \(50 \mathrm{~m} .\) Find the minimum velocity necessary to achieve this range.

Short Answer

Expert verified
The minimum velocity is approximately \(42.14 \, \text{m/s}\).

Step by step solution

01

Understanding the Concepts

We are asked to find the minimum velocity needed for a projectile to reach a certain range when launched at a specific angle. The range of a projectile can be determined using the formula:\[ R = \frac{v^2 \sin(2\theta)}{g} \]where:- \( R \) is the range, 50 meters in this case.- \( v \) is the initial velocity.- \( \theta \) is the launch angle, which is \(8^{\circ}\).- \( g \) is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\).
02

Set Up the Equation

We need to solve for the initial velocity \( v \) using the given range of 50 meters and the angle of \(8^{\circ}\). Substitute these values into the range formula:\[ 50 = \frac{v^2 \sin(16^{\circ})}{9.81} \]
03

Solve for Initial Velocity

First, calculate \( \sin(16^{\circ}) \). Using a calculator, \( \sin(16^{\circ}) \approx 0.276 \).Substitute this value into the equation:\[ 50 = \frac{v^2 \times 0.276}{9.81} \]Rearrange the equation to solve for \( v^2 \):\[ v^2 = \frac{50 \times 9.81}{0.276} \]Calculate this value.
04

Compute the Result

Perform the multiplication:\[ 50 \times 9.81 = 490.5 \]Now divide by the sine value:\[ \frac{490.5}{0.276} \approx 1777.17 \]Take the square root:\[ v \approx \sqrt{1777.17} \approx 42.14 \]Thus, the minimum initial velocity is approximately \(42.14 \, \text{m/s}\).

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Key Concepts

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

Range of a Projectile
The range of a projectile is an important concept in physics. It determines how far a projectile will travel when launched from a specific point. To calculate this distance, we use the formula:\[ R = \frac{v^2 \sin(2\theta)}{g} \]Let's break this formula down:
  • **\( R \)** represents the range of the projectile, or how far it travels horizontally. In this case, we are aiming for a range of 50 meters.
  • **\( v \)** is the initial velocity. This is what we need to find in the problem.
  • **\( \theta \)** is the launch angle relative to the horizontal, which in our exercise is given as 8 degrees.
  • **\( g \)** is the acceleration due to gravity, typically approximated as 9.81 m/s虏 on Earth's surface.
The sine term, \( \sin(2\theta) \), helps adjust for the effect of the launch angle on the range. The formula accounts for both the vertical and horizontal components of velocity. Understanding this helps in calculating how the angle and initial speed affect how far the projectile goes.
Initial Velocity Calculation
Calculating initial velocity is crucial in projectile motion problems, especially when you need a projectile to hit a specific target. To find the initial velocity using the given values, we substitute into the formula for range:\[ 50 = \frac{v^2 \sin(16^{\circ})}{9.81} \]Here, 50 meters is the range we're aiming to achieve. We perform the calculations in small steps:
  • Calculate the sine of 16掳, which results from doubling our 8掳 launch angle. Using a calculator, \( \sin(16^{\circ}) \approx 0.276 \).
  • Next, rearrange the formula to solve for \( v^2 \): \[ v^2 = \frac{50 \times 9.81}{0.276} \]
  • Solve to find \( v^2 = 1777.17 \).
Finally, take the square root of this result to get \( v \approx 42.14 \) m/s. Therefore, the minimum initial velocity required is approximately 42.14 m/s. Paying attention to these calculations ensures accuracy in the projectile's range.
Launch Angle
The launch angle greatly influences the path and range of a projectile. In projectile motion, the angle at which the projectile is launched will determine the shape and reach of its trajectory. Here, our given angle is 8 degrees.The formula for range includes \( \sin(2\theta) \), meaning that it multiplies the given angle by two鈥攆or an 8 degree launch angle, it's calculated using \( 16^{\circ} \).Let's see why each degree matters:
  • Small angles typically result in a flatter trajectory, which decreases the maximum height but increases horizontal travel distance, within certain limits.
  • A launch angle of 8 degrees is relatively shallow, enhancing horizontal range for a given speed.
  • Altering the angle shifts both the vertical and horizontal components of the velocity, demonstrating why optimizing the angle is crucial.
Understanding the role of the launch angle in conjunction with initial velocity helps predict projectile motion more effectively. Adjusting the angle intelligently can help achieve targets more accurately.

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