Question

According to \(Guinness\) \(World\) \(Records\), the longest home run ever measured was hit by Roy "Dizzy" Carlyle in a minor league game. The ball traveled 188 m (618 ft) before landing on the ground outside the ballpark. (a) If the ball's initial velocity was in a direction 45\(^{\circ}\) above the horizontal, what did the initial speed of the ball need to be to produce such a home run if the ball was hit at a point 0.9 m (3.0 ft) above ground level? Ignore air resistance, and assume that the ground was perfectly flat. (b) How far would the ball be above a fence 3.0 m (10 ft) high if the fence was 116 m (380 ft) from home plate?

Short answer

(a) Initial speed is approximately 42.68 m/s. (b) The ball is about 11.98 m above the fence.

Step-by-step solution

Understand the Problem

We need to find the initial velocity of a ball hit at a 45-degree angle above the horizontal that travels 188 m, starting from a height of 0.9 m. Additionally, we want to calculate how far above a 3m high fence, located 116 m away, the ball passes.

Use Projectile Motion Formula

The horizontal range formula for projectile motion is given by: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] where \( R = 188 \, \text{m} \), \( \theta = 45^\circ \), and \( g = 9.81 \, \text{m/s}^2 \). Substitute these values to solve for \( v_0 \).

Calculate Initial Speed

Since \( \theta = 45^\circ \), we have \( \sin(2\theta) = \sin(90^\circ) = 1 \). So the equation becomes: \[ 188 = \frac{v_0^2}{9.81} \]. Solve for \( v_0 \) by rearranging the equation: \[ v_0^2 = 188 \times 9.81 \] \[ v_0 = \sqrt{188 \times 9.81} \]. Calculate to get \( v_0 \).

Calculate for the Fence Height

Using the projectile motion formulas, the vertical position \( y \) as a function of time \( t \) can be given by: \[ y = y_0 + v_{0y}t - \frac{1}{2}gt^2 \] with \( v_{0y} = v_0 \sin(\theta) \). Find \( t \) when horizontal distance \( x = v_{0x}t = 116 \, \text{m} \), where \( v_{0x} = v_0 \cos(\theta) \).

Solve for Time at 116 m

Substitute \( x = 116 \, \text{m} \) into the horizontal motion equation:\[ 116 = v_0 \cos(45^\circ) t \]. Solve for \( t \) using \( v_0 \) from Step 3.

Height Above the Fence

Use the time \( t \) from Step 5 in the equation: \[ y = 0.9 + (v_0\sin(45^\circ))t - \frac{1}{2}(9.81)t^2 \] to find \( y \) at \( x = 116 \, \text{m} \). Then find \( y - 3.0 \, \text{m} \).

Key concepts

Initial Velocity

In the context of projectile motion, the initial velocity is a key component. It's the speed at which an object, like a baseball, is launched into the air. For projectile motion, initial velocity is often represented as a vector, which means it has both a magnitude (how fast) and a direction (which way).
In the exercise, the baseball's initial velocity is crucial to solving how far it will travel and how high it goes over an obstacle. The problem states the baseball was hit at a 45-degree angle, meaning the initial velocity vector is equally divided between horizontal and vertical components.
This initial velocity can be broken down into two parts:

• Horizontal Component ( (v_{0x}) ): This tells us how fast the ball moves horizontally. It is given by (v_0 imes ext{cos}(45^ ext{°})) .
• Vertical Component ( (v_{0y}) ): This tells us how fast the ball moves vertically. It is (v_0 imes ext{sin}(45^ ext{°})) .

The correct initial velocity ensures the ball reaches its target distance and clears obstacles, like the fence mentioned in the exercise.

Horizontal Range Formula

Understanding the horizontal range formula is crucial when studying projectile motion. This formula helps to determine how far a projectile will travel horizontally, given its initial speed and launch angle.
In the problem, the horizontal range ((R)) was 188 meters. The key formula used is:
\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]
Here:

• ((R)) is the horizontal distance.
• ((v_0)) is the initial speed of the ball.
• ((\theta)) is the angle of projection, which was 45 degrees in this case.
• ((g)) is the gravitational constant, approximately 9.81 m/s².

With the launch angle of 45°, (\sin(90°) = 1) , simplifying the formula to (R = \frac{v_0^2}{g}) . This makes calculating the initial speed easier once you know the range. This is a powerful tool for predicting projectile paths.

Vertical Position

The vertical position of a projectile is an important aspect of its trajectory, especially when obstacles are involved. Vertical position ((y)) tells us how high above the ground or any object the projectile is at a given time.
The formula that expresses vertical position during projectile motion is:
\[ y = y_0 + v_{0y}t - \frac{1}{2}gt^2 \]
Here:

• ((y_0)) is the initial height from which the projectile is launched, 0.9 meters in this problem.
• ((v_{0y})) is the initial velocity's vertical component.
• ((t)) is the time since launch.
• ((\frac{1}{2}gt^2)) represents the impact of gravity, pulling the projectile down over time.

In the problem, we calculate how high the ball is as it reaches the fence. This involves determining how long it takes to reach the fence horizontally and then finding its vertical position at that time. This helps in determining if the ball clears the fence, providing insights into both trajectory planning and obstacle avoidance.