A place kicker must kick a football from a point 36 m (about 39 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 24 m/s at an angle of 50� to the horizontal.
The acceleration of gravity is 9.8 m/s2.
To determine if the ball approaches the crossbar while still rising or while falling, what is its vertical velocity at the crossbar?
Answer in units of m/s
To determine the vertical velocity of the ball at the crossbar, we need to analyze the vertical component of the initial velocity.
Given:
Initial velocity (v0) = 24 m/s
Launch angle (θ) = 50 degrees
To find the vertical component of the initial velocity (v0y), we need to use trigonometry. Since the launch angle is given relative to the horizontal, we can use the sine function:
v0y = v0 * sin(θ)
Where:
v0y is the vertical component of the initial velocity.
Plugging in the values:
v0y = 24 m/s * sin(50 degrees)
Calculating this expression:
v0y ≈ 18.30 m/s
Therefore, the vertical velocity of the ball at the crossbar is approximately 18.30 m/s.
To determine the vertical velocity of the ball at the crossbar, we first need to analyze the motion of the ball in the vertical direction.
Given:
Initial speed, v₀ = 24 m/s
Launch angle, α = 50 degrees
Acceleration due to gravity, g = 9.8 m/s²
Vertical distance to the crossbar, d = 3.05 m
We can break down the initial velocity v₀ into its vertical component (v₀y) and horizontal component (v₀x) using trigonometry. The vertical component is given by:
v₀y = v₀ * sin(α)
Substituting the known values:
v₀y = 24 m/s * sin(50 degrees)
v₀y = 18.32 m/s
Now we need to determine if the ball is rising or falling when it reaches the crossbar. We need to calculate the time it takes for the ball to reach the crossbar. We can use the kinematic equation for vertical displacement (d) and time (t):
d = v₀yt - 1/2 * g * t²
Rearranging the equation:
1/2 * g * t² - v₀yt + d = 0
This is a quadratic equation in terms of time (t). Solving for t using the quadratic formula:
t = (-b ± √(b² - 4ac))/2a
Where:
a = 1/2 * g
b = -v₀y
c = d
Substituting the known values:
t = (-(-18.32) ± √((-18.32)² - 4 * 1/2 * 9.8 * 3.05))/(2 * 1/2 * 9.8)
Simplifying the equation:
t = (18.32 ± √(336.1024 - 59.84))/9.8
t = (18.32 ± √(276.2624))/9.8
t = (18.32 ± 16.62)/9.8
We have two solutions for time, t₁ and t₂:
t₁ = (18.32 + 16.62)/9.8
t₁ ≈ 3.42 s
t₂ = (18.32 - 16.62)/9.8
t₂ ≈ 0.17 s
Since time cannot be negative in this context, t = t₂ ≈ 0.17 s.
Now that we have the time it takes for the ball to reach the crossbar, we can determine its vertical velocity at that point.
The vertical velocity (v_y) can be calculated using the equation:
v_y = v₀y - g * t
Substituting the known values:
v_y = 18.32 m/s - 9.8 m/s² * 0.17 s
v_y ≈ 16.61 m/s
Therefore, the vertical velocity of the ball at the crossbar is approximately 16.61 m/s.
Vf^2 = Vo^2 + 2gd,
Vf^2=(18.39)^2 + (-19.6)*3.05 278.41,
Vf = 16.69m/s at the crossbar.
Disregard above solution.
Vo = 24m/s@50deg.
Xo = hor. = 24cos50 = 15.43m/s.
Yo = ver. - 24sin50 = 18.39m/s.
t(up) = (Yf - Yo) / g,
t(up) = (0 - 18.39) / -9.8 = 1.88s.
hmax = (Yf^2 - Yo^2) / 2g,
hmax = (0 - (18.39)^2) / -19.6=17.25m.
Yf^2 = Yo^2 + 2g*(17.25-3.05),
Yf^2 = 0 + 19.6*14.2 = 278.32,
Yf = 16.69 @ the crossbar.