Question

Ask you to solve a variety of problems based on the principles of projectile motion. A baseball player hits a ball at \(100 \mathrm{mph}\), with an initial height of \(3 \mathrm{ft}\) and an angle of elevation of \(20^{\circ}\), at Boston's Fenway Park. The ball flies towards the famed "Green Monster," a wall \(37 f t\) high located \(310 f t\) from home plate. (a) Show that as hit, the ball hits the wall. (b) Show that if the angle of elevation is \(21^{\circ}\), the ball clears the Green Monster.

Short answer

For 20°, the ball hits the wall. For 21°, the ball clears the wall.

Step-by-step solution

Convert Initial Velocity to Feet Per Second

The initial speed of the ball is given as 100 mph. Let's convert this to feet per second (fps) using the conversion factor 1 mph = 1.467 fps. Thus,\[ \text{Velocity in fps} = 100 \text{ mph} \times 1.467 \text{ fps/mph} = 146.7 \text{ fps}. \]

Calculate Horizontal and Vertical Components of Velocity

For an angle of elevation \( \theta \), the horizontal and vertical velocity components \( V_x \) and \( V_y \) are given by:\[ V_x = V \cos(\theta), \quad V_y = V \sin(\theta), \]where \( V = 146.7 \text{ fps} \) and \( \theta = 20^{\circ} \). So,\[ V_x = 146.7 \cos(20^{\circ}), \quad V_y = 146.7 \sin(20^{\circ}). \]

Calculate Time to Reach the Wall

We know that the horizontal distance to the wall is 310 ft. The time \( t \) to reach the wall can be calculated using:\[ t = \frac{\text{distance}}{V_x} = \frac{310}{146.7 \cos(20^{\circ})}. \]

Calculate Vertical Position at the Wall

Using the time from Step 3, calculate the vertical position \( y \) at the wall using the projectile motion equation:\[ y = y_0 + V_y t - \frac{1}{2}gt^2, \]where \( y_0 = 3 \text{ ft} \) is the initial height and \( g = 32.2 \text{ ft/s}^2 \) is the acceleration due to gravity. Substitute values to find \( y \) at \( t = \frac{310}{146.7 \cos(20^{\circ})} \).

Determine if the Ball Hits or Clears the Wall for 20°

Substitute the calculated time in Step 3 back into the vertical position equation in Step 4 to find \( y \). Compare this with the height of the wall (37 ft). If \( y \) is less than or equal to 37 ft, the ball hits the wall.

Repeat for 21° Angle of Elevation

Repeat Steps 2-5 using \( \theta = 21^{\circ} \). Compare the new height \( y \) when the ball reaches the wall to determine if it clears the wall.

Key concepts

Velocity Components

Understanding velocity components is crucial in projectile motion. When an object, like a baseball, is hit at an angle, its velocity can be separated into two parts: horizontal and vertical.
These components describe how fast the object travels in each direction:

• **Horizontal component (\( V_x \)):** This is how fast the object moves sideways. It can be calculated using the formula \( V_x = V \cos(\theta) \), where \( V \) is the initial speed, and \( \theta \) is the angle of elevation.
• **Vertical component (\( V_y \)):** This tells us how fast the object moves upwards or downwards. For calculation, use \( V_y = V \sin(\theta) \).

These components help us predict how high and how far the baseball will go. By analyzing them, you can understand the ball's trajectory and determine if it will hit or clear the wall.

Angle of Elevation

The angle of elevation is an essential part of projectile motion. It is the angle at which the object is launched relative to the ground.
This angle affects both the height the projectile reaches and the distance it travels.

• A **small angle** such as \( 20^{\circ} \) results in a longer horizontal distance but may not achieve great height.
• A **larger angle** like \( 21^{\circ} \) increases the height while slightly reducing horizontal distance.

The angle at which the baseball is hit determines if it climbs over obstacles such as the Green Monster. A change in this angle can be the difference between hitting or clearing a wall. Therefore, analyzing this angle helps in planning precise and effective projectile paths.

Gravity

Gravity plays a central role in projectile motion by constantly pulling the object towards the earth. It affects only the vertical motion.
The gravity on Earth is approximately \( 32.2 \text{ ft/s}^2 \).
Here is what you should note:

• **Consistent force:** Gravity is a constant force that accelerates the object downwards at a steady rate, affecting how high and how long the object will stay in the air.
• **Impact on velocity:** As the object moves upward, gravity slows the vertical velocity. Once peak height is reached, it speeds up the descent.

Next time you see a baseball soar through Fenway Park, remember that gravity is the silent player influencing how the ball moves vertically through its trajectory.

Vertical and Horizontal Motion

The motion of a projectile like a baseball can be broken down into vertical and horizontal components. Both are analyzed individually due to the influence of different forces.
**Vertical motion** is affected by gravity, leading to acceleration downwards. We calculate this motion using:\[ y = y_0 + V_y t - \frac{1}{2}gt^2 \] where \( y_0 \) is the initial height, \( V_y \) is the vertical velocity, and \( g \) is gravity.
**Horizontal motion**, on the other hand, occurs at a constant velocity because no external force influences it horizontally, calculated as:\[ x = V_x t \] where \( V_x \) is the horizontal velocity and \( t \) is time.

• Vertical motion determines the height, affected by both initial vertical speed and gravity.
• Horizontal motion measures distance covered, solely dependent on the initial horizontal velocity.

By analyzing both separately, you can predict how far and how high the ball travels. Combining these insights helps you solve real-life problems like whether a baseball clears a wall or not.