91Ó°ÊÓ

When dealing with projectile motion, understanding the initial velocity components is key. In this problem, the ball is hit with an initial velocity of 40 m/s at an angle of 26° above the horizontal. This initial velocity can be split into two components: horizontal and vertical.
The horizontal component, represented as \( v_{0x} \), is the portion of the initial velocity that acts along the horizontal direction. It is calculated using the trigonometric function cosine as follows:

• \( v_{0x} = v_0 \cdot \cos(26^\circ) \)
• This equates to approximately 36.0 m/s

The vertical component, \( v_{0y} \), is the portion directed vertically. It is found using the sine function:

• \( v_{0y} = v_0 \cdot \sin(26^\circ) \)
• This equals about 17.5 m/s

By knowing these components, we are set to determine how far and high the ball will travel once in motion.

Once the initial velocity components are determined, calculating the time of flight for the projectile becomes straightforward. Here, the horizontal distance traveled by the ball is 110 m, which is the distance to where the fielder stands. Since horizontal motion occurs at a constant speed (neglecting air resistance), we use the formula:

• \( x = v_{0x} \cdot t \)

where \( x \) is the distance and \( v_{0x} \) is the horizontal velocity component. By rearranging, the time \( t \) can be calculated:

• \( t = \frac{x}{v_{0x}} \)
• This results in approximately \( t = \frac{110}{36.0} \approx 3.06 \) seconds

This calculation provides the time it takes for the ball to reach the fielder.

Understanding the vertical position of the ball involves using the kinematic equations. The vertical position of a projectile at any time \( t \) is given by:

• \( y(t) = y_0 + v_{0y} \cdot t - \frac{1}{2}gt^2 \)

In this formula:

• \( y_0 \) is the initial vertical position, 1.2 m in this case
• \( v_{0y} \) is the initial vertical velocity found earlier
• \( g \) is the acceleration due to gravity, and \( t \) is the time calculated to be 3.06 seconds

Using these values in the calculation will yield:

• \( y(3.06) = 1.2 + 17.5 \cdot 3.06 - \frac{1}{2} \cdot 9.8 \cdot (3.06)^2 \approx 8.89 \text{ m} \)

This tells us the height of the ball when it reaches the fielder.

Kinematic equations are instrumental in analyzing projectile motion, allowing the calculation of various trajectory parameters. They combine motion variables like velocity, time, displacement, and acceleration.

• The horizontal motion equation typically uses \( x = v_{0x} \cdot t \)
• Vertical motion often uses \( y(t) = y_0 + v_{0y} \cdot t - \frac{1}{2}gt^2 \)

Each component operates independently, with gravity affecting vertical motion while horizontal motion continues uniformly due to the absence of horizontal forces in ideal conditions. Applying these equations appropriately will provide insights into factors such as displacement, air time, and impact velocities, critical in understanding both theoretical and real-world applications of projectile motion.