Question

Use the model for projectile motion, assuming there is no air resistance. A baseball, hit 3 feet above the ground, leaves the bat at an angle of \(45^{\circ}\) and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise?

Short answer

The initial speed of the baseball is given by the equation \(v_i = x / (t \cos(\theta))\) and the maximum height reached by the baseball is given by the equation \(y_{\textrm{max}} = v_i \sin(\theta) t - \frac{1}{2}g t^2\)

Step-by-step solution

Identify the equations of motion for the projectile

The primary equations of motion for a projectile under gravity without air resistance are \(y = v_i \sin(\theta) t - \frac {1}{2} g t^2\) for the vertical motion and \(x = v_i \cos(\theta) t\) for the horizontal motion, where \(v_i\) is the initial velocity, \(\theta\) is the launch angle, \(g\) is the acceleration due to gravity (approximately \(32.2 \, ft/s^2\) on the surface of the Earth), \(t\) is the time, and \(x\) and \(y\) denote the horizontal and vertical displacements.

Solve for the initial speed

At the time when the ball is caught, the vertical position \(y\) is equal to the initial position, which is 3 feet. Therefore, from the equation for horizontal motion \(x = v_i \cos(\theta)t\), solve for \(v_i\) by substituting the values \(x = 300 ft\), \(\theta = 45^{\circ} = \pi/4\) rad, and \(t\), which can be obtained from \(y = 3 ft\). After solving, obtain \(v_i = x/(t \cos(\theta))\).

Calculate the maximum height

The maximum height of the projectile is reached when the vertical velocity becomes zero i.e. \(v_i \sin(\theta) - g t = 0\). Solving for \(t = t_{\textrm{max}}\), substitute into \(y = y_{\textrm{max}} = v_i \sin(\theta) t - \frac{1}{2}g t^2\), to obtain the maximum height.

Key concepts

Equations of Motion

The equations of motion for projectile motion are the backbone of analyzing the path of an object, like a baseball, as it moves through the air. To solve problems involving projectile motion without air resistance, we utilize two fundamental equations.

The first equation describes vertical motion: \[y = v_i \sin(\theta)t - \frac{1}{2}gt^2\]Here, \(y\) represents the vertical position of the projectile, \(v_i\) is the initial velocity, \(\theta\) is the launch angle, \(g\) is the acceleration due to gravity (which is about 32.2 ft/s^2 on Earth), and \(t\) is the time elapsed. This equation incorporates gravity's role in pulling the projectile back down towards the ground.

The second equation focuses on horizontal motion:\[x = v_i \cos(\theta)t\]In this case, \(x\) is the horizontal distance from the origin. It's based on the assumption that in the absence of air resistance, the horizontal component of a projectile's velocity remains constant. The initial speed can be uncovered by blending these equations with known values—including the distance to the catcher and the fact the baseball lands at the same vertical position from which it was hit.

Initial Velocity

Understanding initial velocity (\(v_i\)) is crucial for solving projectile motion problems. It's the speed at which the baseball is moving just as it leaves the bat and begins its arc. With both the angle and the point of landing known, we can determine the initial velocity using the horizontal motion equation mentioned before.

To calculate \(v_i\), we first need to know the time of flight, which we get by analyzing the vertical motion. Since the ball lands at the same height it was hit, we set the final vertical position to be equal to the initial vertical position. By solving for time using vertical motion, we can then find the initial velocity by restructuring the horizontal equation as:\[v_i = \frac{x}{t \cos(\theta)}\]Here, \(x\) is the distance to the outfielder (300 feet), \(\theta\) is the launch angle which at \(45^\circ\) or \(\pi/4\) radians, makes calculations easier because \(\cos(\theta)\) equals \(\sin(\theta)\), and \(t\) is the time. Once the time is known from the vertical motion, the initial velocity is quickly found, revealing how swiftly the ball was traveling when it was first hit.

Maximum Height

The maximum height of a projectile is the peak of its trajectory. To determine this, we need to find the point at which the vertical component of the projectile's velocity is zero because this is the moment it stops rising and begins to fall.

The vertical component of the velocity can be expressed as:\[v_i \sin(\theta) - g t = 0\]To find the time at which the maximum height is reached (\(t_{\textrm{max}}\)), we set the above expression to zero and solve for \(t\). Substituting \(t_{\textrm{max}}\) back into the vertical motion equation, we get the maximum height (\(y_{\textrm{max}}\)):\[y_{\textrm{max}} = v_i \sin(\theta) t - \frac{1}{2}g t^2\]In this case, once the initial velocity (\(v_i\)) is known from previous calculations, it all boils down to plugging that value along with the time of reaching the maximum height into the vertical motion equation. This gives us the highest point the baseball reaches before gravity pulls it back down towards the hands of the outfielder. Determining the maximum height is not just a common academic exercise; it's also a valuable piece of information for sports strategists and can change the way players are trained and games are played.