91Ó°ÊÓ

The projectile range refers to the horizontal distance a projectile covers during its flight. For a projectile launched from a height and returning to the same height, this distance depends on the initial velocity, launch angle, and the acceleration due to gravity.
The formula to calculate this range is given by:\[ R = \frac{v_0^2 \sin 2\theta}{g} \]where:

• \( R \) is the range,
• \( v_0 \) is the initial velocity,
• \( \theta \) is the launch angle,
• \( g \) is the acceleration due to gravity, typically \(9.81 \mathrm{~m/s^2}\).

In this exercise, the ball has a range of 107 meters, which means that's the horizontal distance it travels from the bat until it returns to its original height. Understanding projectile range helps in determining whether the ball can clear obstacles like fences.

A projectile's motion can be broken down into horizontal and vertical components. This bifurcation helps in addressing the complex movement separately.
Horizontal Component:The horizontal component of velocity remains constant throughout the flight, owing to the absence of horizontal forces (ignoring air resistance). It's calculated as:\[ v_{0x} = v_0 \cos \theta \]Vertical Component:The vertical component, however, is affected by gravity, which accelerates the object downward. It's calculated as:\[ v_{0y} = v_0 \sin \theta \]In our scenario, where the ball is launched at a 45-degree angle, both components have the same magnitude because \( \sin 45^{\circ} = \cos 45^{\circ} \).
Knowing these components is vital for calculating the individual movements in each direction and predicting the projectile's path.

Determining initial velocity allows us to delve into the future motion of a projectile. It's a critical calculation for comprehending how far and high a projectile will travel.
Given the formula for range and the necessary variables:\[ 107 = \frac{v_0^2 \sin 90^{\circ}}{9.81} \]We solve for \( v_0 \) as follows:

• Since \( \sin 90^{\circ} = 1 \), it simplifies our calculation.
• \( v_0^2 = 107 \times 9.81 \)
• \( v_0^2 = 1049.67 \)
• \( v_0 = \sqrt{1049.67} \approx 32.41 \mathrm{~m/s} \)

This initial velocity of 32.41 m/s is crucial for further computations, such as determining time of flight, maximum height, and whether the ball will clear obstacles during its path.

The calculation of trajectory height involves determining the highest point that a projectile reaches in its flight path. This situation often involves assessing whether a projectile can pass over an obstacle.
Using the equation for vertical motion:\[ y = y_0 + v_{0y} t - \frac{1}{2} g t^2 \]For a projectile at time \( t = 4.26 \mathrm{~s} \), the prerequisites are:

• \( y_0 = 1.22 \mathrm{~m} \) (the initial height of the ball),
• \( v_{0y} = 22.91 \mathrm{~m/s} \) (the vertical component of initial velocity).

Plugging these into the formula:\[ y = 1.22 + 22.91 \cdot 4.26 - \frac{1}{2} \cdot 9.81 \cdot (4.26)^2 \]\[ y \approx 9.7 \mathrm{~m} \]This height calculation tells us the ball reaches a height of 9.7 meters at the distance of 97.5 meters away. This insight confirms whether the ball clears the 7.32-meter fence by assessing the differential height.