Question

A kicker kicks a football off the ground when he is located 30 meters away (measured horizontally) from the goalposts. He kicks it at an angle of 30◦ above the horizontal. With what minimum speed (in m/s) does he need to kick the ball in order to get it over the crossbar (3 meters off the ground)?

Answer 1

Answer 2

To determine the minimum speed required for the ball to clear the crossbar, we can use the principles of projectile motion. Let's break down the problem and solve it step by step:

1. The horizontal distance between the kicker and the goalposts is 30 meters.

2. The ball needs to clear a vertical distance of 3 meters to reach the crossbar.

3. The angle of projection is 30 degrees above the horizontal.

In projectile motion, we can analyze the vertical and horizontal components of the ball's velocity separately.

Vertical Component: The vertical motion of the ball is influenced by gravity. We know that the initial vertical velocity (Viy) is determined by the initial speed (Vi) and the angle of projection (θ). Using trigonometry, we can find Viy:

Viy = Vi * sin(θ)

Horizontal Component: The horizontal motion of the ball is unaffected by gravity. The horizontal velocity (Vix) remains constant throughout the motion. We can find Vix using the initial speed (Vi) and the angle of projection (θ):

Vix = Vi * cos(θ)

To clear a vertical distance of 3 meters, the time of flight (t) is the same for the upward and downward paths. We can calculate t using the vertical equation of motion:

Δy = Viy * t + (1/2) * g * t^2

Since the ball goes up and comes back down to the same vertical level, Δy = 3 meters. Also, g is the acceleration due to gravity (approximately 9.8 m/s^2).

3 = Vi * sin(θ) * t - (1/2) * g * t^2

To find the minimum speed required, we need to determine the time of flight. We can solve the quadratic equation for t:

(1/2) * g * t^2 - Vi * sin(θ) * t + 3 = 0

Once we find the time of flight (t), we can calculate the minimum initial speed (Vi) using the horizontal equation of motion:

Δx = Vix * t

Since Δx is the horizontal distance of 30 meters, we can solve for Vi:

30 = Vi * cos(θ) * t

Now, let's substitute the known values into the equations and solve for Vi:

θ = 30 degrees Δy = 3 meters Δx = 30 meters g = 9.8 m/s^2

First, let's find t by solving the quadratic equation:

(1/2) * (9.8) * t^2 - Vi * sin(30) * t + 3 = 0

Next, solve for t. Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

where a = (1/2) * (9.8), b = -Vi * sin(30), and c = 3.

Simplifying the equation:

t = (-(-Vi * sin(30)) ± √((-Vi * sin(30))^2 - 4 * (1/2) * (9.8) * 3)) / (2 * (1/2) * (9.8))

Now, substitute the values and solve for t:

t = (Vi * sin(30) ± √((Vi * sin(30))^2 - 2.94)) / 4.9

Since the time of flight is positive, we take the positive value:

t = (Vi * sin(30) + √((Vi * sin(30))^2 - 2.94)) / 4.