A baseball was hit at 45m/s at an angle of 45 degrees above the horizontal. How long did it remain in the air? How far did it travel horizontally?
To find the time of flight and horizontal distance traveled by the baseball, we can use the equations of motion for projectile motion.
First, we need to split the initial velocity of the baseball into its horizontal and vertical components. The horizontal component of velocity (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to the influence of gravity.
Vx = V * cos(theta)
Vy = V * sin(theta)
Given:
Initial velocity (V) = 45 m/s
Launch angle (theta) = 45 degrees
Plugging in these values, we can calculate:
Vx = 45 * cos(45)
Vx = 45 * √(2)/2
Vx ≈ 31.82 m/s
To find the time of flight (T), we can use the following equation:
T = (2 * Vy) / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Vy = V * sin(theta)
Vy = 45 * sin(45)
Vy = 45 * √(2)/2
Vy ≈ 31.82 m/s
T = (2 * 31.82) / 9.8
T ≈ 6.48 seconds
So, the baseball remained in the air for approximately 6.48 seconds.
To find the horizontal distance traveled (D), we can use the equation:
D = Vx * T
D = 31.82 * 6.48
D ≈ 206.28 meters
Therefore, the baseball traveled approximately 206.28 meters horizontally.