A football kicker can give the ball an initial speed of 25m/s. What are the greatest and least elevation angles at which he can kick the ball to score a field goal from a point 50m in front of goalposts whose horizontal bar is 3.44m above the ground?
31 to 63 degrees above the horizontal.
To find the greatest and least elevation angles, we can make use of the following equation:
range = ((v^2 * sin(2θ)) / g)
where:
- v is the initial speed of the ball
- θ is the elevation angle
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Given:
- v = 25 m/s
- range = 50 m
We can rearrange the equation to solve for θ:
θ = (1/2) * arcsin((g * range) / v^2)
Now, let's calculate the greatest and least elevation angles.
First, let's calculate the greatest elevation angle:
θ_greatest = (1/2) * arcsin((9.8 * 50) / 25^2)
θ_greatest ≈ 16.3 degrees
Now, let's calculate the least elevation angle:
θ_least = (1/2) * arcsin((9.8 * 50) / 25^2)
θ_least ≈ -16.3 degrees (measured below the horizontal)
However, since an elevation angle cannot be negative in this context, we can discard the negative value. Therefore, the least elevation angle is 16.3 degrees (measured above the horizontal).
So, the greatest elevation angle at which the kicker can kick the ball to score a field goal is approximately 16.3 degrees, and the least elevation angle is also approximately 16.3 degrees.
To find the greatest and least elevation angles at which the football kicker can kick the ball to score a field goal, we can use projectile motion equations.
Let's break down the problem into components:
1. Initial speed (v0) = 25 m/s
2. Horizontal distance (range) = 50 m
3. Height of the goalposts (h) = 3.44 m
We need to find the angle at which the ball should be kicked to reach the goalposts.
Let's start by finding the greatest elevation angle:
1. The horizontal component of velocity (v0x) remains constant throughout the entire motion since there is no acceleration in the horizontal direction. Therefore, v0x = v0 * cos(theta), where theta is the angle of elevation.
Using the given initial speed of 25 m/s, we can calculate the horizontal component of velocity:
v0x = 25 * cos(theta)
2. The vertical component of velocity (v0y) changes due to the acceleration due to gravity. We can use the equation v0y = v0 * sin(theta) - g * t, where g is the acceleration due to gravity (9.8 m/s²) and t is the time of flight.
3. The time of flight can be calculated using the equation t = 2 * v0y / g.
4. To reach the goalposts, the vertical displacement (h) should be equal to the height of the goalposts.
Using the equation h = v0y * t - (1/2) * g * t^2, we can substitute the values:
3.44 = (v0 * sin(theta) - g * t) * t - (1/2) * g * t^2
Now, we can solve this equation to find the greatest elevation angle.
Similarly, we can find the least elevation angle by considering the maximum possible horizontal distance, which is 50 meters.
Solve the equation h = (v0 * sin(theta) - g * t) * t - (1/2) * g * t^2 for theta, taking into account the range of 50 m.
Using these equations and solving them either analytically or numerically, you can find the greatest and least elevation angles at which the football kicker can kick the ball to score a field goal.