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

Short Answer

Expert verified
The horizontal distance traveled by the shot is approximately 21.84 meters.

Step by step solution

01

Identify Given Values

We know that the initial speed of the shot, \(v_0 = 14.4\, \text{m/s}\), the angle of projection \(\theta = 34.0^\circ\), the initial height of the shot \(h = 2.10\, \text{m}\), and the acceleration due to gravity \(g = 9.81\, \text{m/s}^2\).
02

Calculate Horizontal and Vertical Components of Velocity

Using trigonometry, calculate the horizontal component \(v_{0x} = v_0 \cdot \cos(\theta)\) and the vertical component \(v_{0y} = v_0 \cdot \sin(\theta)\). Calculate these as follows:\[ v_{0x} = 14.4 \cdot \cos(34.0^\circ) \approx 12.0\, \text{m/s} \]\[ v_{0y} = 14.4 \cdot \sin(34.0^\circ) \approx 8.1\, \text{m/s} \]
03

Calculate Time in Air Using Vertical Motion Equation

We use the vertical motion formula \(y = v_{0y} \cdot t + \frac{1}{2} \cdot a \cdot t^2\), where \(y\) is the vertical displacement (final height - initial height = 0 - 2.10), and \(a = -g\). Substitute the known values:\[-2.10 = 8.1 \cdot t - 0.5 \cdot 9.81 \cdot t^2\].This is a quadratic equation \(-4.905t^2 + 8.1t - 2.10 = 0\). Solve this quadratic for \(t\) using the quadratic formula, \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -4.905\), \(b = 8.1\), and \(c = -2.10\).
04

Solve the Quadratic Equation for Time

Using the quadratic formula:\[ t = \frac{-8.1 \pm \sqrt{(8.1)^2 - 4(-4.905)(-2.10)}}{2(-4.905)} \]Simplify to find the positive root for \(t\). After computation, \(t \approx 1.82\, \text{s}\).
05

Calculate Horizontal Distance

The horizontal distance \(x\) can be found using \(x = v_{0x} \cdot t\). Substitute the values:\[ x = 12.0 \times 1.82 \approx 21.84\, \text{m} \]
06

Conclusion

The shot travels approximately 21.84 meters horizontally.

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Key Concepts

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

Kinematics
Kinematics is the branch of mechanics concerned with the motion of objects without considering the forces that cause the motion. In projectile motion, we observe how an object moves when it is launched into the air at an angle. This motion can be broken down into horizontal and vertical components, which are independent of each other. The horizontal component deals with movement across a plane, unaffected by gravity, while the vertical component includes the effects of gravity.

In our shot-putter problem, to look into kinematics, we separate the initial throw into two components: the horizontal velocity and the vertical velocity. We assume that the object moves under its initial velocity and gravity's constant acceleration. This separation helps in using kinematic equations to solve for other quantities such as time in the air or distance traveled. The key equations relate initial velocity, time, acceleration (gravity), and displacement. These elements help describe the shot put's flight in precise terms with measurements.

Sometimes, finding these components helps in determining how an object behaves when launched at different angles or velocities, making kinematics useful in various real-world scenarios.
Trigonometry in Physics
Trigonometry in physics often involves breaking down vectors into horizontal and vertical components using trigonometric functions like sine and cosine. In projectile motion, we have to deal with angles and velocities, which makes trigonometry particularly handy.

For example, when our shot-putter launches the shot at an angle of 34 degrees with an initial speed of 14.4 meters per second, trigonometry allows us to determine how much of that speed aims horizontally and how much vertically. We use the formula:
  • Horizontal component: \( v_{0x} = v_0 \cdot \cos(\theta) \)
  • Vertical component: \( v_{0y} = v_0 \cdot \sin(\theta) \)
By plugging in our values, we find the horizontal and vertical components of our shot's initial velocity. The horizontal component keeps the shot moving across the field, while the vertical component fights against gravity to give the shot its arching path.

Understanding these concepts helps us predict where the shot will land and how far it will travel, making trigonometry a crucial tool in physics calculations.
Quadratic Equations in Physics
Quadratic equations in physics emerge when dealing with projectile motion, especially to solve for time or displacement. In our exercise, the vertical motion of the shot leads us to a quadratic equation due to its parabolic trajectory.

We start with the vertical motion equation: \[ y = v_{0y} \cdot t + \frac{1}{2} \cdot a \cdot t^2 \]where \( y \) is the vertical displacement. After inserting our values, we see that calculating time requires solving a quadratic equation: \[ -4.905t^2 + 8.1t - 2.10 = 0 \]Quadratic equations generally take the form \( ax^2 + bx + c = 0 \) and can be solved by applying the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Our criteria focus on selecting the positive root as time cannot be negative. Solving this equation gives us the time the shot is in the air.

These equations are fundamental in physics to find how objects move through space when under the influence of gravity and initial velocities.

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Most popular questions from this chapter

(II) A child, who is 45 m from the bank of a river, is being carried helplessly downstream by the river's swift current of 1.0 m/s. As the child passes a lifeguard on the river's bank, the lifeguard starts swimming in a straight line (Fig. 3-46) until she reaches the child at a point downstream. If the lifeguard can swim at a speed of 2.0 m/s relative to the water, how long does it take her to reach the child? How far downstream does the lifeguard intercept the child?

(a) A long jumper leaves the ground at 45\(^\circ\) above the horizontal and lands 8.0 m away. What is her "takeoff" speed \(v_0\)? (\(b\)) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m vertically below. If she long jumps from the edge of the left bank at 45\(^\circ\) with the speed calculated in (\(a\)), how long, or short, of the opposite bank will she land (Fig. 3-50)?

(II) A grasshopper hops along a level road. On each hop, the grasshopper launches itself at angle \(\theta_0 =\) 45\(^\circ\) and achieves a range \(R =\) 0.80 m. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between hops is negligible.

(II) A projectile is fired with an initial speed of 36.6 m/s at an angle of 42.2\(^\circ\) above the horizontal on a long flat firing range. Determine (\(a\)) the maximum height reached by the projectile, (\(b\)) the total time in the air, (\(c\)) the total horizontal distance covered (that is, the range), and (\(d\)) the speed of the projectile 1.50 s after firing.

(I) A car is driven 225 km west and then 98 km southwest (45\(^\circ\)). What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.

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