Question

A batter hits a pitched ball at a height \(4.0 \mathrm{ft}\) above the ground so that its angle of projection is \(45^{\circ}\) and its horizontal range is \(350 \mathrm{ft}\). The ball travels down the left field line where a 24 -fthigh fence is located \(320 \mathrm{ft}\) from home plate. Will the ball clear the fence? If so, by how much?

Short answer

The short answer will be if the ball can cross the fence or not based on the calculated height \( y \) and the actual height of the fence. Also, by how much does the ball clear the fence will be the difference between \( y \) and the height of the fence.

Step-by-step solution

Calculating the initial velocity

Firstly, we can calculate the initial velocity (magnitude of the velocity vector when the ball was hit). We know that the range of a projectile fired at an angle \( \theta \) is given by the equation: \( R = \frac{v_{0}^{2} \cdot \sin(2\theta)}{g} \) Here, \( R = 350 \, ft \), \( \theta = 45^{\circ} \) and \( g = 32.2 \, ft/s^{2} \) (acceleration due to gravity). By plugging these values into the formula, we can solve for \( v_{0} \).

Calculating the time to reach the fence

The horizontal component of the velocity at which the ball was hit can be calculated using the relation \( v_{0x} = v_{0} \cdot \cos(\theta) \). By using this component, we can find out the time \( t \) it takes for the ball to reach the fence \( 320 \, ft \) away by using the formula \( x = v_{0x} \cdot t \) and solve for \( t \).

Calculating the Height

Now we calculate the vertical position of the ball at this time \( t \). Equation for the vertical position of the ball, starting from an initial height \( h_{0} \), is \( y = -\frac{1}{2}gt^2 + v_{0y}t + h_{0} \). Here, we can find \( v_{0y} \) using the relation \( v_{0y} = v_{0} \cdot \sin(\theta) \). Substituting the values of \( g, v_{0y}, t, h_{0} \), we can solve for \( y \).

Key concepts

Initial Velocity

Understanding the initial velocity of a projectile is crucial for analyzing its motion. In the context of our baseball problem, the initial velocity is the speed at which the batter hits the ball. It's represented by the symbol \( v_0 \) and is a vector quantity, meaning it has both magnitude and direction.

In projectile motion, the initial velocity can be broken down into two components: horizontal (\( v_{0x} \)) and vertical (\( v_{0y} \)). For a projectile launched at angle \( \theta \), the horizontal component is calculated by \( v_{0} \cdot \cos(\theta) \) and the vertical component by \( v_{0} \cdot \sin(\theta) \). These components are important, as they are individually responsible for the distance traveled horizontally and the maximum height reached by the projectile.

Using the range formula, \( R = \frac{v_{0}^{2} \sin(2\theta)}{g} \), and the given range and projection angle, we can solve for the initial velocity, which provides a foundation for subsequent calculations in our exercise.

Range of a Projectile

The range of a projectile is the horizontal distance it travels before landing. It's one of the most observed outcomes when analyzing projectile motion. The formula for the range, when the initial and final vertical positions are the same, which is true in many cases like in our baseball problem, is \( R = \frac{v_{0}^{2} \sin(2\theta)}{g} \), where \( R \) is the range, \( v_{0} \) is the initial velocity, \( \theta \) is the launch angle, and \( g \) is the acceleration due to gravity.

In a situation where the projectile is launched and lands at different heights, the calculation requires adjusting for this difference. The range is crucial, as it dictates if the projectile will reach a desired target or not — in our case, whether the baseball will clear the fence.

Vertical Position

The vertical position of a projectile at any point in time is key to predicting and understanding its flight path. In projectile motion, this is often where gravity plays its most visible role. The formula for a projectile's vertical position \( y \) at time \( t \) is \( y = -\frac{1}{2}gt^2 + v_{0y}t + h_{0} \), where \( g \) is the acceleration due to gravity, \( v_{0y} \) is the initial vertical velocity, and \( h_{0} \) is the initial height from which the projectile is launched.

In a realistic scenario like our baseball problem, the ball does not start from ground level; it starts from a height of 4 feet. Therefore, we need to include this initial height in our calculations to determine the vertical position of the ball accurately when it reaches the fence. This tells us whether the ball goes over the fence or not.

Time of Flight

Time of flight in projectile motion refers to the total time a projectile is airborne. This duration is significant because it impacts the distance traveled horizontally and the final vertical position of the projectile. To find the time it takes for a projectile to reach a certain horizontal position, we can use the formula \( x = v_{0x} \cdot t \) where \( x \) is the horizontal distance, \( v_{0x} \) is the horizontal component of the initial velocity, and \( t \) is the time.

In our baseball problem, to calculate the time of flight until the ball reaches the fence, we use the time formula and solve for \( t \), given the known horizontal component of the initial velocity and the horizontal distance to the fence. Estimating the time of flight accurately is pivotal, as it syncs up with the vertical motion to give a full picture of the projectile's trajectory.