A place kicker must kick a football from a point 33.0 m (about 36 yd) from the goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 48◦ to the horizontal.
To determine if the ball clears the crossbar, what is its height with respect to the crossbar when it reaches the plane of the crossbar? The acceleration of gravity is 9.81 m/s2 .
Answer in units of m
As you know, the trajectory of a body kicked with velocity v at angle θ is
y = x tanθ - g/(2 (v cosθ)^2) x^2
For your ball, that is
y(x) = 1.11x - 9.81/(2(20*0.67)^2) x^2
= 1.11x - 0.0273x^2
So, now check to see whether y(33) >= 3.05
To determine the height of the ball with respect to the crossbar when it reaches the plane of the crossbar, we can use the kinematic equations of motion.
First, let's break down the initial velocity of the ball into its horizontal and vertical components.
The horizontal component of the initial velocity is given by v₀ₓ = v₀ * cos(θ), where v₀ is the initial speed (20.0 m/s) and θ is the angle (48 degrees).
v₀ₓ = 20.0 m/s * cos(48 degrees) = 13.0 m/s
The vertical component of the initial velocity is given by v₀ᵧ = v₀ * sin(θ).
v₀ᵧ = 20.0 m/s * sin(48 degrees) = 15.0 m/s
Now, let's find the time it takes for the ball to reach the plane of the crossbar. We can use the equation:
y = y₀ + v₀ᵧ * t + (1/2) * (-g) * t²
Since the final vertical position is equal to the height of the crossbar (3.05 m) and the initial vertical position (y₀) is 0 m (assuming the ground is y = 0), we can solve for t.
3.05 m = 0 m + 15.0 m/s * t + (1/2) * (-9.81 m/s²) * t²
Rearranging this equation and using the quadratic formula, we get:
(-4.905 m/s²) * t² + 15.0 m/s * t - 3.05 m = 0
Solving this quadratic equation, we find two possible values for t: t₁ = 0.321 s and t₂ = 2.51 s. Since time cannot be negative in this context, we ignore t₁.
So, it takes approximately t = 2.51 seconds for the ball to reach the plane of the crossbar.
Now, let's find the height of the ball (y-coordinate) when it reaches the crossbar by using the equation:
y = y₀ + v₀ᵧ * t + (1/2) * (-g) * t²
Substituting the values, we have:
y = 0 m + 15.0 m/s * (2.51 s) + (1/2) * (-9.81 m/s²) * (2.51 s)²
Simplifying the equation, we get:
y = 37.59 m + (-31.96 m) ≈ 5.63 m
Therefore, when the ball reaches the plane of the crossbar, its height with respect to the crossbar is approximately 5.63 meters.