There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle?
Theta = arcantan(height/range)
To determine the minimum angle above the ground, let's consider the concept of projectile motion.
Projectile motion is the motion of an object that is launched into the air and moves along a curved path under the influence of gravity. In this case, the object is a ball being kicked towards a bar.
To clear the bar, the ball needs to reach a certain height at the point where it passes over the bar. This can be achieved by giving the ball an initial velocity and launch angle.
The equation for the height of a projectile at any given time is given by:
h = (vi^2 * sin^2θ) / (2g)
Where:
h = maximum height reached by the projectile
vi = initial velocity of the projectile
θ = launch angle
g = acceleration due to gravity (usually taken as 9.8 m/s^2)
In the given scenario, we want to find the minimum launch angle such that the ball cannot clear the bar, regardless of its initial velocity.
To determine this angle, we can set the height (h) equal to the height of the bar. Let's assume the height of the bar is denoted by 'Bar_h'.
Bar_h = (vi^2 * sin^2θ) / (2g)
Rearranging the equation for the launch angle (θ) gives us:
sin^2θ = (2g * Bar_h) / vi^2
Since sin^2θ cannot exceed 1, the maximum value on the right side of the equation is 1. Therefore:
1 = (2g * Bar_h) / vi^2
Now, we can solve for the minimum launch angle by isolating θ:
θ = arcsin(sqrt((vi^2) / (2g * Bar_h)))
After substituting the values of vi, g, and Bar_h, you can calculate the minimum launch angle required.