Samma Sosa hit a ball at 40 m/s at a 22 degree angle. Find the following:
Time the ball was in the air
Horizontal distance
Maximum height of the ball
To find the time the ball was in the air, horizontal distance, and maximum height of the ball, we can use the basic equations of projectile motion.
1. Time the ball was in the air:
The time of flight can be calculated using the vertical motion equation:
t = (2 * v * sinθ) / g
Where:
t = time of flight
v = initial velocity of the ball
θ = launch angle
g = acceleration due to gravity (approximately 9.8 m/s²)
Substituting the given values:
v = 40 m/s
θ = 22°
t = (2 * 40 * sin22°) / 9.8
t ≈ 4.20 seconds
Therefore, the ball was in the air for approximately 4.20 seconds.
2. Horizontal distance:
The horizontal distance covered by the ball can be found using the horizontal motion equation:
Horizontal distance = v * cosθ * t
Substituting the given values:
v = 40 m/s
θ = 22°
t = 4.20 seconds
Horizontal distance = 40 * cos22° * 4.20
Horizontal distance ≈ 142.6 meters
Therefore, the ball traveled approximately 142.6 meters horizontally.
3. Maximum height of the ball:
To find the maximum height of the ball, we can use the vertical motion equation:
Maximum height = (v² * sin²θ) / (2 * g)
Substituting the given values:
v = 40 m/s
θ = 22°
g = 9.8 m/s²
Maximum height = (40² * sin²22°) / (2 * 9.8)
Maximum height ≈ 28.7 meters
Therefore, the maximum height of the ball is approximately 28.7 meters.