A football kicker lines up to kick a game winning kick. The goalpost is 30m away and the ball must clear the 3.3m high cross-bar. If the kick leaves his foot at an angle of 35 degrees, how fast must the kick be to be good?
d = Vo^2*sin(2A)/g = 30.
Vo^2*sin70/9.8 = 30,
Vo^2 * 0.0959 = 30,
Vo = 17.7 m/s.
To find out how fast the kick must be to clear the crossbar, we can break down the initial velocity into vertical and horizontal components.
1. Find the vertical component of the initial velocity:
The ball must clear a height of 3.3m, and the initial angle is 35 degrees.
Vertical velocity (Vy) = V * sin(θ)
Where V is the initial velocity and θ is the angle of elevation.
Vy = V * sin(35°)
2. Find the horizontal component of the initial velocity:
The distance to the goalpost is 30m, and the initial angle is 35 degrees.
Horizontal velocity (Vx) = V * cos(θ)
Where V is the initial velocity and θ is the angle of elevation.
Vx = V * cos(35°)
3. The time it takes for the ball to reach the goalpost horizontally is the same as the time it takes for the ball to reach its peak vertically.
Using the formula time (t) = Vertical velocity (Vy) / acceleration due to gravity (g), we can find the time it takes for the ball to reach its peak.
t = (V * sin(35°)) / 9.8
4. The vertical distance the ball travels is given by the equation:
Vertical distance = Vy * t + (0.5) * (-9.8) * t^2
However, at the top of its trajectory, the vertical velocity becomes 0, so the equation can be simplified to:
Vertical distance = (0.5) * (-9.8) * t^2
The vertical distance must be greater than the height of the crossbar (3.3m):
(0.5) * (-9.8) * t^2 > 3.3
5. Solve the inequality for t:
(0.5) * (-9.8) * t^2 > 3.3
(-4.9) * t^2 > 3.3
t^2 < (3.3 / -4.9)
t^2 < -0.673
Since time cannot be negative, there is no solution for t. This means that the kick cannot clear the crossbar.
Therefore, it is not possible to determine how fast the kick must be to be good as it cannot clear the crossbar with an initial angle of 35 degrees.
To determine the necessary speed for the kick to be successful, we can use the principles of projectile motion. The velocity of the ball can be divided into its horizontal and vertical components.
First, let's find the vertical component of the velocity. We know that the ball must clear a 3.3m high cross-bar, so the maximum height of the ball's trajectory will be half of that, which is 1.65m. We can use the equation for the vertical motion of a projectile:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
y = vertical distance (1.65m)
y0 = initial vertical position (0m)
v0y = initial vertical velocity (unknown)
t = time of flight
g = acceleration due to gravity (-9.8 m/s^2)
Since the ball reaches the maximum height halfway through the total time of flight, we can use the formula:
t = (2 * v0y) / g
Substituting the values in the equation:
1.65 = 0 + (v0y * (2 * v0y / g)) - (1/2)(9.8)(2 * v0y / g)^2
Simplifying the equation:
1.65 = (2v0y^2) / 9.8
From this equation, we can solve for v0y, the initial vertical velocity.
Next, let's find the horizontal component of the velocity. The horizontal distance the ball needs to travel is 30m, and the time of flight is the same for both horizontal and vertical motion. Therefore, we can use the formula:
x = v0x * t
Where:
x = horizontal distance (30m)
v0x= initial horizontal velocity (unknown)
t = time of flight (which we previously found)
Simplifying the equation:
30 = v0x * (2v0y / g)
Now, we have two equations. We can substitute the value of v0y from the first equation into the second equation and solve for v0x.
Once we have both the vertical and horizontal components of the velocity (v0y and v0x), we can find the magnitude of the velocity (v0) using the Pythagorean theorem:
v0 = sqrt(v0x^2 + v0y^2)
This magnitude of velocity should be the minimum speed required for the kick to be successful.
Keep in mind that in this explanation, we assumed certain conditions like no air resistance and neglecting factors like wind.
Horizontal: d=Vh*time, or a) timeinair=30/(V*cos35)
vertical: Hf=hi+vi*t- gt^2/2 or b) 3.3=0+V*sin35:*t-4.9 t^2
Now put in these two equations for t, put time in air , and then solve for V