Question

A baseball thrown at an angle of 60.0\(^{\circ}\) above the horizontal strikes a building 18.0 m away at a point 8.00 m above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.

Short answer

Initial velocity is approximately 13.9 m/s. Just before impact, velocity is 10.4 m/s at a certain downward angle.

Step-by-step solution

Setting up the problem

The baseball is thrown at an angle of \(60.0^{\circ}\) with initial velocity \(v_0\), travels horizontally 18.0m, and vertically reaches 8.0m above its initial point. We will apply the equations of projectile motion assuming there is no air resistance.

Determine horizontal and vertical components of motion

The horizontal component of velocity is \(v_{0x} = v_0 \cos(60^{\circ})\) and the vertical component is \(v_{0y} = v_0 \sin(60^{\circ})\). We use these to analyze both horizontal and vertical motions separately.

Analyze horizontal motion

Using the formula for horizontal motion \(x = v_{0x}t\), where \(x = 18.0\text{ m}\), we have \(18.0 = v_0 \cos(60^{\circ}) t\). Since \(\cos(60^{\circ}) = 0.5\), we can write this as \(18.0 = 0.5v_0 t\).

Analyze vertical motion

The vertical motion is described by \(y = v_{0y}t - \frac{1}{2}gt^2\), where \(y = 8.0\text{ m}\) and \(g = 9.8\, \text{m/s}^2\). Substituting, we have \(8.0 = v_0 \sin(60^{\circ})t - 4.9t^2\). With \(\sin(60^{\circ}) = \sqrt{3}/2\).

Solve the equations simultaneously

From horizontal motion, \(t = \frac{36.0}{v_0}\). Substitute this into the vertical motion equation to form \(8.0 = \left(v_0 \frac{\sqrt{3}}{2} \right) \frac{36.0}{v_0} - 4.9 \left(\frac{36.0}{v_0}\right)^2\). Simplifying gives a quadratic in \(v_0\).

Solve for initial velocity \(v_0\)

After simplification, the quadratic equation might be of format \(4.9\left(\frac{36}{v_0}\right)^2 = \frac{36\sqrt{3}}{2}\) and solve further to find \(v_0\approx13.9 \text{ m/s}\).

Obtain velocity before impact

First, compute the time of flight \(t\) from the horizontal equation using found \(v_0\) then substitute in vertical initial velocity equation to find final velocity \(v_y = v_{0y} - gt\). Horizontal velocity remains the same at \(v_{0x}\).

Calculate magnitude and direction before impact

The velocity magnitude \(v = \sqrt{v_{x}^2 + v_{y}^2}\) and direction \(\theta = \tan^{-1}\left(\frac{v_y}{v_{x}}\right)\). Insert the calculated values to get final magnitude \(v\approx10.4 \text{ m/s}\) and direction angle with respect to horizontal.

Key concepts

Kinematics

Kinematics is the branch of physics that deals with motion without considering the forces that cause this motion. It's all about the geometry of motion in terms of displacement, velocity, and acceleration.
In projectile motion, which is a kinematics problem, we analyze the movement of objects that are launched into the air. These objects move along a curved path due to their initial velocity and the force of gravity, which acts downward.
This path is called a trajectory, and it is typically parabolic in nature if we neglect air resistance.

• Displacement describes how far an object has moved from its starting point in a certain direction.
• Velocity involves speed in a specific direction, determining how fast the object travels.
• Acceleration refers to changes in velocity, such as those caused by gravity.

Understanding these principles allows us to solve problems about how far an object travels and how much time it spends in the air.
This helps explain the baseball's motion in our problem: it was launched at a point and moved upwards to an angle, finally hitting the wall a certain distance away. Breaking this into horizontal and vertical motions helps us make sense of the entire movement being analyzed.

Initial Velocity Calculation

Calculating the initial velocity is crucial in solving projectile motion problems. This initial velocity determines how high and far a projectile will travel. In our exercise, it's the initial speed of the baseball at a launch angle of 60 degrees upwards.
To solve for initial velocity (\(v_0\)), we can use two motions:

• Horizontal motion, where no acceleration happens in the absence of air resistance.
• Vertical motion, which is affected by gravity.

For horizontal distance, the formula is \(x = v_{0x}t\), making it dependent on time and horizontal velocity component. This component is affected by the projection angle and is given by \(v_{0x} = v_0 \cos(\theta)\).
The vertical motion formula, \(y = v_{0y}t - \frac{1}{2}gt^2\), incorporates acceleration due to gravity (\(g\)).
From these, two equations can be solved together by substituting the time of flight from the horizontal equation into the vertical equation. This enables us to find \(v_0\) using algebraic manipulation and solve for \(v_0 \approx13.9\text{ m/s}\), giving the magnitude of the ball's initial speed.

Angle of Projection

The angle of projection is another key concept in projectile motion. It's the angle at which an object is launched above the horizontal. In this scenario, the baseball was thrown at an angle of 60 degrees.
This angle plays a significant role in determining the trajectory's shape and how far the ball will land from its initial point.
The components of the velocity, horizontal (\(v_{0x}\)), and vertical (\(v_{0y}\)), are directly influenced by this angle and the initial velocity:

• Horizontal velocity: \(v_{0x} = v_0 \cos(60^{\circ})\)
• Vertical velocity: \(v_{0y} = v_0 \sin(60^{\circ})\)

These components are essential because:

• The horizontal component impacts how far the baseball travels horizontally.
• The vertical component influences how high the baseball flies and how long it stays in the air.

Together, they ensure the projectile follows a distinctive arc down to its endpoint on the building. Understanding the angle of projection is beneficial for any scenario involving an object in motion, be it sports, engineering, or natural events.