A rugby player kicks a ball from a ground aiming to get the ball over rhe horizontal bar and betwween the vertical posts. the ball is 50 m away from goal line, directly infront of a goal, and the initial speed of the ball is 25 m/s. The horizontal bar is 3.5 from the ground. Between what two initial elevation angles must the ball be kicked to get over the bar?
To find the initial elevation angles required for the ball to clear the horizontal bar, we can use the projectile motion equations.
Let's consider the following variables:
- v = initial speed of the ball = 25 m/s
- g = acceleration due to gravity = 9.8 m/s^2
- d = distance from the goal line to the ball = 50 m
- h = height of the horizontal bar = 3.5 m
We need to find the two angles (θ1 and θ2) at which the ball must be kicked to clear the bar. We'll assume symmetry, so θ1 will be the lower angle and θ2 will be the higher angle.
The horizontal distance covered by the ball can be calculated using the equation:
d = (v^2 * sin(2θ)) / g
Since we know the horizontal distance, we can rearrange the above equation to solve for sin(2θ):
sin(2θ) = (d * g) / v^2
Next, we can solve for 2θ:
2θ = sin^(-1)((d * g) / v^2)
Now, we can solve for θ1 and θ2:
θ1 = (1/2) * (sin^(-1)((d * g) / v^2))
θ2 = 90 - θ1
Substituting the given values:
θ1 = (1/2) * (sin^(-1)((50 * 9.8) / 25^2))
θ2 = 90 - θ1
Calculating the values gives:
θ1 ≈ 18.4 degrees
θ2 ≈ 71.6 degrees
Therefore, the ball must be kicked at an initial elevation angle between approximately 18.4 degrees and 71.6 degrees to clear the horizontal bar.
To determine the range of initial elevation angles required for the ball to clear the horizontal bar, we need to consider the projectile motion of the ball. We can use the equations of motion to solve for the required angles.
Given:
- Initial speed of the ball (u) = 25 m/s
- Distance from the ball to the goal line (range) = 50 m
- Height of the horizontal bar (height) = 3.5 m
Step 1: Find the time of flight (t)
The time of flight is the total time it takes for the ball to travel from the kicker to the goal line. We can use the equation:
range = u * cos(theta) * t
where theta is the angle of projection measured from the horizontal axis.
Solving for t, we get:
t = range / (u * cos(theta))
Step 2: Find the maximum height (H)
The maximum height achieved by the ball can be calculated using the equation:
H = (u^2 * sin^2(theta)) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Step 3: Check if the ball clears the bar
To clear the bar, the maximum height (H) should be greater than the height of the horizontal bar (height).
If H > height, then the ball clears the bar.
Step 4: Solve for the angles
We need to find the range of initial elevation angles theta that satisfy both the time of flight and clearing the bar conditions.
By substituting the expression for t from Step 1 into the equation for H, we can write:
H = (u^2 * sin^2(theta)) / (2 * g)
Simplifying, we can rewrite this as:
sin^2(theta) = (2 * H * g) / u^2
Taking the square root of both sides, we get:
sin(theta) = sqrt((2 * H * g) / u^2)
And finally, solving for theta:
theta = arcsin(sqrt((2 * H * g) / u^2))
Now, substitute the values of H, g, and u into the equation to get the range of initial elevation angles.