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A projectile is launched at ground level with an initial speed of \(50.0 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the \(x\) and \(y\) distances from where the projectile was launched to where it lands?

Short Answer

Expert verified
The projectile lands at a horizontal distance (\(x\)) of 129.9 meters and a vertical distance (\(y\)) of 34.3 meters from the launch point.

Step by step solution

01

Find the horizontal and vertical components of the initial velocity

Separate the initial velocity into horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components using trigonometry. For the horizontal component, use the formula \[v_{0x} = v_0 \cos(\theta)\] and for the vertical component, use \[v_{0y} = v_0 \sin(\theta)\] where \(v_0\) is the initial velocity and \(\theta\) is the launch angle.
02

Calculate the horizontal distance

Use the horizontal component of the initial velocity to find the horizontal distance (\(x\)) using the formula \[x = v_{0x} \times t\] since there is no horizontal acceleration.
03

Calculate the vertical distance

Calculate the vertical distance (\(y\)) using the kinematic equation for motion under constant acceleration: \[y = v_{0y} \times t + \frac{1}{2} g t^2\] where \(g\) is the acceleration due to gravity (negative in this case since it's in the downward direction) and \(t\) is the time of flight. However, as the projectile lands above the ground, we consider the vertical acceleration as positive.
04

Solve for the horizontal and vertical distances

Plug in the values for the horizontal and vertical components obtained from Step 1 into the formulas provided in Steps 2 and 3, along with the given values for time and acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)).

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Key Concepts

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

Kinematics in Two Dimensions
In physics, the study of motion is known as kinematics, and dealing with objects moving in two dimensions brings an additional level of complexity compared to one-dimensional motion.
Understanding projectile motion involves analyzing the movement both horizontally and vertically. Since these two components of motion are perpendicular to each other, they can be treated independently. An object in projectile motion will move at a constant velocity horizontally (if we neglect air resistance) while simultaneously accelerating vertically due to gravity.

To comprehensively analyze two-dimensional motion, we break down the initial velocity into horizontal and vertical components using trigonometric functions. Distances traveled in each direction can then be calculated using kinematic equations, which describe the relationships between displacement, velocity, acceleration, and time.
Initial Velocity Components
Understanding the initial velocity components is crucial for solving problems in projectile motion. When a projectile is launched, its initial velocity is often given at an angle. This velocity vector can be split into two perpendicular components: horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)).

The horizontal velocity component, \(v_{0x}\), dictates how far the projectile will travel in the horizontal direction and is calculated using the cosine of the launch angle. It remains constant throughout the flight if we neglect air resistance. The vertical velocity component, \(v_{0y}\), initially dictates how high the projectile will go and is calculated using the sine of the launch angle. Unlike \(v_{0x}\), \(v_{0y}\) changes over time because of the gravitational force pulling the projectile down.
Trigonometric Functions in Physics
Trigonometric functions such as sine and cosine are indispensable tools in physics, especially when resolving vector quantities into their components. The initial velocity of a projectile is a vector that has both magnitude and direction.

By using the sine function in the context of projectile motion, we're able to determine the vertical component of initial velocity, \(v_{0y} = v_0 \times \text{sin}(\theta)\), relative to the angle of projection, \theta\. Similarly, the cosine function helps us find the horizontal component, \(v_{0x} = v_0 \times \text{cos}(\theta)\). These functions are vital because they allow us to transform the problem from two dimensions back into two separate one-dimensional problems, which are simpler to solve.
Kinematic Equations
Kinematic equations are the backbone of solving problems in projectile motion. They relate the initial and final velocities, acceleration, time, and displacement. Since projectile motion can be considered in terms of horizontal and vertical motions, we use different kinematic equations for each direction.

In the horizontal direction, the kinematic equation simplifies to distance equals velocity times time, as there is no acceleration \(x = v_{0x} \times t\). In the vertical direction, the equation needs to account for gravitational acceleration \(y = v_{0y} \times t + \frac{1}{2} g t^2\), which dramatically affects the projectile's motion.

Understanding these equations is paramount for predicting the projectile's behavior, such as how high it will climb, how long it will stay in the air, and how far it will travel before hitting the ground or, as in our problem, striking a target at a certain height.

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