A ball is thown at 20m/s at an angle of 30 degrees from the horizontal level. What maximum height can the ball reach?
To find the maximum height the ball can reach, we need to analyze the vertical motion of the ball. We can break the initial velocity of the ball into its vertical and horizontal components.
Given:
Initial velocity (v0) = 20 m/s
Angle of projection (θ) = 30 degrees
First, we find the vertical component of the initial velocity (v0y) using trigonometry:
v0y = v0 * sin(θ)
v0y = 20 * sin(30)
v0y ≈ 10 m/s
Now, we can use the vertical motion equation to find the maximum height (h_max):
h_max = (v0y^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
h_max = (10^2) / (2 * 9.8)
h_max ≈ 5.1 meters
Therefore, the maximum height the ball can reach is approximately 5.1 meters.