Problem 58 - College Physics

Question

A stone thrown off a bridge \(20 \mathrm{~m}\) above a river has an initial velocity of \(12 \mathrm{~m} / \mathrm{s}\) at an angle of \(45^{\circ}\) above the horizontal (vFig. 3.33). (a) What is the range of the stone? (b) At what velocity does the stone strike the water?

Step-by-Step Solution

Answer

The range is approximately 24.29 meters, and the impact velocity is about 21.33 m/s.

Step 1: Initial Velocity Components

First, decompose the initial velocity into horizontal and vertical components. The initial velocity, \( v_0 = 12 \mathrm{~m/s} \), is at an angle of \( 45^{\circ} \). Using trigonometry, the horizontal component is \( v_{0x} = v_0 \cos(45^{\circ}) = 12 \times \frac{\sqrt{2}}{2} \approx 8.49 \mathrm{~m/s} \). The vertical component is \( v_{0y} = v_0 \sin(45^{\circ}) = 12 \times \frac{\sqrt{2}}{2} \approx 8.49 \mathrm{~m/s} \).

Step 2: Time of Flight

Next, calculate the time of flight using the vertical motion equations. The equation for vertical motion is \( y = v_{0y}t + \frac{1}{2}at^2 \), where \( y = -20 \mathrm{~m} \) (since it falls 20 meters), and \( a = -9.81 \mathrm{~m/s^2} \). Using the quadratic formula on \( 0 = 8.49t - 4.905t^2 - 20 \), we find that the positive time of impact is \( t \approx 2.86 \mathrm{~s} \).

Step 3: Calculate the Range

Now, calculate the range using the horizontal velocity and the time of flight. The range \( R \) is given by \( R = v_{0x} \times t \). Substituting the values, \( R = 8.49 \times 2.86 \approx 24.29 \mathrm{~m} \).

Step 4: Final Velocity Components

To find the velocity at impact, determine the final vertical and horizontal velocity components. The horizontal velocity remains constant as \( v_{0x} = 8.49 \mathrm{~m/s} \). For the vertical component, use \( v_{y} = v_{0y} + at \). Substituting in values, \( v_y = 8.49 + (-9.81)(2.86) \approx -19.56 \mathrm{~m/s} \).

Step 5: Resultant Impact Velocity

Calculate the magnitude of the resultant velocity vector using the horizontal and vertical components at impact: \( v = \sqrt{v_{x}^2 + v_{y}^2} \). Substitute the values, \( v = \sqrt{8.49^2 + (-19.56)^2} \approx 21.33 \mathrm{~m/s} \).

Key Concepts

Initial Velocity

When discussing projectile motion, the term **initial velocity** refers to the speed and direction at which an object is first launched. In our exercise, a stone is thrown from a bridge with an initial velocity of \(12 \mathrm{~m/s}\) at a \(45^{\circ}\) angle above the horizontal. Initial velocity affects both the object's trajectory and the distance it will travel. The object's velocity at launch is a combination of speed and direction.

This makes the initial velocity a vector quantity, meaning it has both magnitude and direction. To analyze projectile motion further, this vector must be broken down into horizontal and vertical components. This breakdown allows us to use mathematical formulas to predict the stone's path and landing point.

Horizontal Component

The **horizontal component** of velocity in projectile motion is a key factor, as it influences how far the object will travel horizontally before hitting the ground. Since our initial velocity is \(12\mathrm{~m/s}\) at an angle of \(45^{\circ}\), we can calculate the horizontal component using trigonometry.

For this purpose, we use the cosine function:

• \(v_{0x} = v_0 \cos(\theta)\)

Where \(v_0\) is the initial speed and \(\theta\) is the launch angle. Therefore, the horizontal component is:

• \(v_{0x} = 12 \times \frac{\sqrt{2}}{2} = 8.49 \mathrm{~m/s}\)

The horizontal velocity remains constant during the object's flight because there is no horizontal acceleration in ideal projectile motion without air resistance.

Vertical Component

In projectile motion, the **vertical component** of velocity is crucial because it determines the height and flight duration of the object. For the stone, the vertical component at launch can also be determined using trigonometry.

Here, we apply the sine function:

• \(v_{0y} = v_0 \sin(\theta)\)

Where \(v_0\) is the initial velocity and \(\theta\) is the angle of launch. Therefore, the vertical component is:

• \(v_{0y} = 12 \times \frac{\sqrt{2}}{2} = 8.49 \mathrm{~m/s}\)

The vertical component is affected by gravity, which acts downwards at \(-9.81 \mathrm{~m/s^2}\). This gravitational force will alter the stone's vertical velocity throughout its trajectory, causing it to eventually strike the water.

Quadratic Formula

The **quadratic formula** is a fundamental tool in algebra that is used to solve quadratic equations, which are polynomial equations of the second degree. In this problem, the quadratic formula helps us determine the time of flight by solving the vertical motion equation.

The vertical position \(y(t)\) as a function of time is given by:

• \(-20 = 8.49t - 4.905t^2\)

This represents a quadratic equation in the form \(at^2 + bt + c = 0\), where:

• \(a = -4.905\)
• \(b = 8.49\)
• \(c = -20\)

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the positive root provides the time of impact:

• \(t \approx 2.86 \mathrm{~s}\)

Finding the positive time ensures we're calculating when the stone actually hits the water after being thrown.