A Baseball player leads off the game and hits along home run. The ball leaves the bat at an angle of 30° with a velocity of 70 m/s
•Determine the horizontal velocity (Vx) of the ball.
•Determine the vertical velocity (Vy) of the ball
• Determine the time when the ball will reach the highest point
•Determine the maximum range of the ball
•Determine the maximum height of the ball
Vx = 70 m/s * cos(30º)
Vy = 70 m/s * sin(30º)
time to max height ... Tmax = Vy / g
max range = Vx * 2 Tmax
max height ... Hmax = Tmax * Vy / 2
To solve these problems, we can use the principles of projectile motion. Projectile motion can be analyzed by separating the motion into horizontal and vertical components.
1. Horizontal Velocity (Vx):
The horizontal velocity remains constant throughout the ball's trajectory, as there are no external horizontal forces affecting it. Therefore, the horizontal velocity (Vx) of the ball remains the same as the initial velocity (70 m/s) since there is no acceleration in the horizontal direction. Thus, Vx = 70 m/s.
2. Vertical Velocity (Vy):
The vertical velocity of the ball changes due to the acceleration caused by gravity. To calculate the vertical velocity (Vy), we need to consider the initial velocity (V0) and the angle (θ) of the ball's trajectory. The vertical velocity can be determined using the following formula:
Vy = V0 * sin(θ)
Given that the initial velocity (V0) is 70 m/s and the angle (θ) is 30°, we can calculate the vertical velocity (Vy):
Vy = 70 m/s * sin(30°)
Vy = 35 m/s (rounded to the nearest whole number)
3. Time to Reach Highest Point:
To determine the time when the ball will reach the highest point, we can use the formula:
t = Vy / g
Where:
t is the time,
Vy is the vertical velocity, and
g is the acceleration due to gravity (approximately 9.8 m/s²)
Substituting the values:
t = 35 m/s / 9.8 m/s²
t ≈ 3.57 seconds (rounded to two decimal places)
4. Maximum Range:
The maximum range is the horizontal distance traveled by the ball before hitting the ground. To calculate the maximum range, we can use the formula:
Range = Vx * t
Substituting the values:
Range = 70 m/s * 3.57 seconds
Range ≈ 249.9 meters (rounded to one decimal place)
5. Maximum Height:
The maximum height can be determined by calculating the vertical displacement (Δy), which is the change in the vertical position of the ball.
Δy = Vy² / (2 * g)
Substituting the values:
Δy = (35 m/s)² / (2 * 9.8 m/s²)
Δy ≈ 62.3 meters (rounded to one decimal place)
Therefore, the answers to the questions are:
• Horizontal velocity (Vx) of the ball: 70 m/s
• Vertical velocity (Vy) of the ball: 35 m/s
• Time when the ball reaches the highest point: Approximately 3.57 seconds
• Maximum range of the ball: Approximately 249.9 meters
• Maximum height of the ball: Approximately 62.3 meters
To solve the given problem, we need to consider the following information:
Angle of launch (θ) = 30°
Initial velocity (V) = 70 m/s
1. Determine the horizontal velocity (Vx) of the ball:
The horizontal velocity remains constant throughout the projectile motion. It is given by the formula Vx = V * cos(θ), where Vx is the horizontal velocity and θ is the angle of launch. Plugging in the values:
Vx = 70 m/s * cos(30°)
Vx ≈ 60.62 m/s
2. Determine the vertical velocity (Vy) of the ball:
The vertical velocity changes due to the effect of gravity. Initially, the vertical velocity is given by Vy = V * sin(θ), where Vy is the vertical velocity and θ is the angle of launch. Plugging in the values:
Vy = 70 m/s * sin(30°)
Vy ≈ 35 m/s
3. Determine the time when the ball will reach the highest point:
The time taken to reach the highest point can be found using the formula t = Vy / g, where t is the time, Vy is the vertical velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values:
t = 35 m/s / 9.8 m/s^2
t ≈ 3.57 seconds
4. Determine the maximum range of the ball:
The maximum range of the ball is the horizontal distance traveled. It can be calculated using the formula R = Vx * t, where R is the range, Vx is the horizontal velocity, and t is the time taken to reach the highest point. Plugging in the values:
R = 60.62 m/s * 3.57 seconds
R ≈ 216.65 meters
5. Determine the maximum height of the ball:
To find the maximum height, we can use the formula h = (Vy^2) / (2g), where h is the maximum height, Vy is the vertical velocity, and g is the acceleration due to gravity. Plugging in the values:
h = (35 m/s)^2 / (2 * 9.8 m/s^2)
h ≈ 61.73 meters
Therefore, the answers are:
- The horizontal velocity (Vx) of the ball is approximately 60.62 m/s.
- The vertical velocity (Vy) of the ball is approximately 35 m/s.
- The time when the ball will reach the highest point is approximately 3.57 seconds.
- The maximum range of the ball is approximately 216.65 meters.
- The maximum height of the ball is approximately 61.73 meters.