A place-kicker kicks a football at an angle of è = 41.5° above the horizontal axis, as the figure below shows. The initial speed of the ball is v0 = 26 m/s. Also assume that the ball is kicked on the moon instead of on the earth. The acceleration due to gravity on the moon has a magnitude of 1.62 m/s2
Nice. What is the question?
To solve this problem, we can use the formulas of projectile motion. Let's break down the given information:
Angle of kick (θ): θ = 41.5°
Initial speed of the ball (v0): v0 = 26 m/s
Acceleration due to gravity on the moon (gmoon): gmoon = 1.62 m/s²
1. Determine the components of initial velocity:
The initial velocity of the ball can be broken down into its horizontal and vertical components.
v0x = v0 * cos(θ)
v0x = 26 m/s * cos(41.5°)
v0x ≈ 19.89 m/s (rounded to two decimal places)
v0y = v0 * sin(θ)
v0y = 26 m/s * sin(41.5°)
v0y ≈ 16.88 m/s (rounded to two decimal places)
2. Determine the time of flight:
The time it takes for the ball to reach the ground can be found using the vertical component of the motion. We will ignore air resistance in this calculation.
The formula for the time of flight (t) is:
t = (2 * v0y) / gmoon
t = (2 * 16.88 m/s) / 1.62 m/s²
t ≈ 20.83 s (rounded to two decimal places)
3. Determine the horizontal range:
The horizontal distance the ball travels can be found using the horizontal component of the motion.
The formula for the horizontal range (R) is:
R = v0x * t
R = 19.89 m/s * 20.83 s
R ≈ 414.18 m (rounded to two decimal places)
So, on the moon, the football will travel approximately 414.18 meters in distance.