Problem 19 - Thomas Calculus

Question

Travel time A projectile is fired at a speed of 840 \(\mathrm{m} / \mathrm{sec}\) at an angle of \(60^{\circ} .\) How long will it take to get 21 \(\mathrm{km}\) downrange?

Step-by-Step Solution

Answer

The projectile will take 50 seconds to travel 21 km downrange.

Step 1: Identify the formula and known values

To find the time of flight to a specific horizontal distance, use the formula for range: \( R = \frac{v^2 \sin(2\theta)}{g} \), where \( R \) is the range, \( v \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity \( 9.8 \, \text{m/s}^2 \). We know \( v = 840 \, \text{m/s} \), \( \theta = 60^\circ \), and \( R = 21000 \, \text{m} \) (converted from 21 km).

Step 2: Calculate horizontal component of velocity

The horizontal component of the velocity, \( v_x \), is given by \( v \cos(\theta) \). Calculate it: \( v_x = 840 \cos(60^\circ) = 840 \times 0.5 = 420 \, \text{m/s} \).

Step 3: Use horizontal motion formula to find time

The horizontal range \( R \) can also be expressed as \( R = v_x t \). Solve for \( t \): \( t = \frac{R}{v_x} = \frac{21000}{420} = 50 \, \text{seconds} \).

Key Concepts

Range Formula

When dealing with projectile motion, the **Range Formula** is a crucial mathematical tool. It helps compute how far a projectile will travel horizontally. The formula is given by \( R = \frac{v^2 \sin(2\theta)}{g} \), where:

• \( R \) is the range or horizontal distance covered.
• \( v \) stands for the initial velocity of the projectile.
• \( \theta \) represents the launch angle with respect to the horizontal axis.
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

To find the range, you need both the speed and the angle at which the projectile is launched. This formula assumes ideal conditions: no air resistance and that the initial and final elevations are equal. It's valuable in physics for predicting projectile paths.

Horizontal Component of Velocity

The **Horizontal Component of Velocity** is key to understanding a projectile's flight. In projectile motion, velocity is split into two components: horizontal and vertical. For the horizontal part, you calculate it using the formula \( v_x = v \cos(\theta) \), where:

• \( v_x \) is the horizontal velocity component.
• \( v \) is the initial velocity of the projectile.
• \( \theta \) is the launch angle.

The horizontal velocity component remains constant since no horizontal forces are acting on the projectile in ideal conditions. This simplifies analysis because the range of the projectile, \( R \), is directly proportional to \( v_x \). So, if you want to find how fast the projectile moves along the horizontal plane, calculating \( v_x \) is essential.

Time of Flight Calculation

Understanding the **Time of Flight Calculation** allows you to determine how long a projectile remains in the air. For projectiles launched at an angle, the time of flight depends on both the horizontal velocity and the distance covered. Use the relationship \( R = v_x t \) to find time \( t \):

• \( R \) is the horizontal range.
• \( v_x \) is the horizontal component of velocity.
• \( t \) is the time of flight.

Rearranging this equation, you find \( t = \frac{R}{v_x} \). It's a straightforward calculation when you know \( R \) and \( v_x \). This formula assumes that the projectile lands at the same vertical level from which it was launched, and it's ideal for non-complex environments where external forces like air resistance are neglected. By understanding these concepts, you can accurately predict how long a projectile will stay airborne.