Problem 35 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A shot-putter throws the shot (mass \(=7.3 \mathrm{kg}\) ) with an initial spced of 14.4 \(\mathrm{m} / \mathrm{s}\) at a \(34.0^{\circ}\) angle to the borizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.10 \(\mathrm{m}\) above the ground.

Step-by-Step Solution

Answer

The shot travels approximately 22.8 meters horizontally.

Step 1: Break Down Initial Velocities

First, decompose the initial speed into horizontal and vertical components. Given the initial speed \(v = 14.4\,\mathrm{m/s}\) and the angle \(\theta = 34.0^\circ\):- The horizontal velocity \(v_x = v \cdot \cos(\theta) = 14.4 \cdot \cos(34.0^\circ)\)- The vertical velocity \(v_y = v \cdot \sin(\theta) = 14.4 \cdot \sin(34.0^\circ)\)Calculating these gives:\(v_x \approx 12.0\,\mathrm{m/s}\)\(v_y \approx 8.1\,\mathrm{m/s}\)

Step 2: Calculate Time of Flight

Using the vertical motion equation to find the time the shot is in the air. The vertical equation under gravity is:\[ y = v_y \cdot t + \frac{1}{2}gt^2 \]Given initial height \(y_0 = 2.10\,\mathrm{m}\), and final height \(y = 0\,\mathrm{m}\) at ground level:\[ 0 = 2.10 + (8.1 \cdot t) - \frac{1}{2} \cdot 9.8 \cdot t^2 \]This simplifies to the quadratic equation:\[ 0 = -4.9t^2 + 8.1t + 2.1 \]Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), solve for \(t\).\[ t = \frac{-8.1 \pm \sqrt{8.1^2 + 4 \cdot 4.9 \cdot 2.1}}{9.8} \approx 1.9 \text{ seconds} \]

Step 3: Calculate Horizontal Distance

With the time of flight and horizontal velocity, calculate the horizontal distance traveled. Using the formula:\[ x = v_x \cdot t \]where \(v_x = 12.0\,\mathrm{m/s}\) and \(t = 1.9\,\mathrm{s}\):\[ x = 12.0 \cdot 1.9 \approx 22.8\,\mathrm{m} \]

Key Concepts

Horizontal Velocity

In projectile motion, horizontal velocity is a crucial component influencing how far an object can travel horizontally. When an object is launched, its initial velocity can be split into horizontal and vertical components, especially if it's launched at an angle. The formula for finding horizontal velocity (\(v_x\)) is:

• \(v_x = v \cdot \cos(\theta)\)

where \(v\) is the initial speed, and \(\theta\) is the angle of launch.

Horizontal velocity remains constant throughout the motion as no acceleration affects it horizontally (ignoring air resistance). In our exercise, the shot-putt had an initial speed of 14.4 m/s. With the angle given as 34.0 degrees, the horizontal component was calculated as approximately 12.0 m/s. This consistent velocity across the time of flight helps in computing the total horizontal distance traveled.

Time of Flight

Time of flight is the duration an object spends in motion from the moment it's launched until it hits the ground. In projectile motion, this depends on vertical motion, as vertical velocity (\(v_y\)) and gravity influence how long the projectile remains airborne.

For an object starting at an initial height, the time of flight can be derived using the formula:

• \(0 = y_0 + v_y \cdot t - \frac{1}{2}gt^2\)

Here, \(y_0\) is 2.10 meters, and the object must reach ground level (\(y = 0\)). The vertical component of initial velocity (\(v_y\)) is 8.1 m/s.

By rearranging the equation and applying the quadratic formula, the time calculated for the shot-put to be in the air is approximately 1.9 seconds. This parameter is vital for further calculations, like determining the horizontal distance.

Quadratic Equation

The quadratic equation comes into play in projectile motion to solve for variables such as the time of flight, when vertical motion is not straightforward. When breaking down vertical motion using gravitational effects, you derive an equation in the form of:

• \(0 = at^2 + bt + c\)

For projectile motion, \(a\) is typically \(-\frac{1}{2}g\) (where \(g\) is 9.8 m/s²), \(b\) is the initial vertical velocity, and \(c\) is the initial height.

In our exercise, the quadratic equation formed was:

• \(0 = -4.9t^2 + 8.1t + 2.1\)

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), you can solve time in flight. Solving this led to a practical time, showing how crucial the quadratic equation is in solving real-world physics problems where direct computation isn't feasible due to multiple factors at play.