To win the game, a place kicker must kick a
football from a point 54 m (59.0544 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 26 m/s at
an angle of 33.8
◦
from the horizontal.
The acceleration of gravity is 9.8 m/s
2
.
By how much vertical distance does the ball
clear the crossbar?
Answer in units of m.
Why did the football player bring a ladder to the game? Because he wanted to kick the ball over the crossbar, of course!
Let's calculate how much the ball clears the crossbar. We can break down the motion of the ball into its horizontal and vertical components.
The initial vertical velocity is given by V₀y = V₀ * sin(θ), where V₀ is the initial speed (26 m/s) and θ is the angle (33.8 degrees).
V₀y = 26 m/s * sin(33.8°) = 14 m/s.
Using the equation h = V₀y * t + (1/2) * g * t², where h is the vertical distance, g is the acceleration due to gravity (-9.8 m/s²), and t is the time of flight, which we need to find.
We know that the horizontal distance is given by x = V₀x * t, where V₀x is the initial horizontal velocity. V₀x = V₀ * cos(θ) = 26 m/s * cos(33.8°) = 21.65 m/s.
The time of flight, t, can be found using x = V₀x * t. Rearranging the equation, t = x / V₀x = 54 m / 21.65 m/s ≈ 2.5 s.
Now, let's calculate the vertical distance the ball clears the crossbar using h = V₀y * t + (1/2) * g * t².
h = 14 m/s * 2.5 s + (1/2) * (-9.8 m/s²) * (2.5 s)² ≈ 35.5 m.
So, the ball clears the crossbar by approximately 35.5 meters vertically.
To find the vertical distance the ball clears the crossbar, we need to calculate the maximum height the ball reaches above the crossbar.
Step 1: Split the initial velocity into horizontal and vertical components.
The horizontal component of the initial velocity is given by:
Vx = V * cos(θ)
Vx = 26 m/s * cos(33.8°)
Vx = 21.62 m/s
The vertical component of the initial velocity is given by:
Vy = V * sin(θ)
Vy = 26 m/s * sin(33.8°)
Vy = 14.07 m/s
Step 2: Calculate the time taken for the ball to reach the maximum height.
We know that the ball reaches its maximum height when Vy = 0.
Using the equation:
Vy = Voy - g * t
0 = 14.07 m/s - 9.8 m/s^2 * t
9.8 m/s^2 * t = 14.07 m/s
t = 1.44 s
Step 3: Calculate the maximum height reached by the ball.
Using the equation:
Δy = Voy * t - (1/2) * g * t^2
Δy = 14.07 m/s * 1.44 s - (1/2) * 9.8 m/s^2 * (1.44 s)^2
Δy = 20.258 m - 10.59 m
Δy = 9.668 m
Therefore, the ball clears the crossbar by approximately 9.668 meters.
To find the vertical distance by which the ball clears the crossbar, we can break down the vertical and horizontal components of the projectile motion.
First, let's find the time it takes for the ball to reach its peak height. We can use the vertical motion equation:
y = y0 + v0y * t - (1/2) * g * t^2
where:
- y is the vertical displacement (the height above the ground)
- y0 is the initial vertical position (0 in this case, since the ball starts from the ground)
- v0y is the initial vertical velocity component (v0 * sin(angle))
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time
Since the ball reaches its peak when it has zero vertical velocity, we can set v0y = 0 and solve for t:
0 = v0y - g * t
v0y = g * t
Now, let's find the time it takes for the ball to reach the ground after being kicked. Again, we can use the vertical motion equation, but this time with y = -3.05 m (negative because we count downward from the crossbar):
-3.05 = y0 + v0y * t - (1/2) * g * t^2
-3.05 = 0 + (g * t) * t - (1/2) * g * t^2
Now, let's solve this quadratic equation for t:
(1/2) * g * t^2 - (g * t) * t - 3.05 = 0
We can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
where:
a = (1/2) * g = 4.9 m/s^2
b = -(g) = -9.8 m/s^2
c = -3.05 m
Now, we have two values for t: one for when the ball is kicked and one for when it reaches the ground. We only want the positive value of t, representing the time it takes for the ball to reach its peak and then fall to the ground.
Once we have this value of t, we can substitute it back into the vertical motion equation to find the vertical distance by which the ball clears the crossbar, y:
y = y0 + v0y * t - (1/2) * g * t^2
Finally, we can calculate the numerical value of y in meters to determine how much vertical distance the ball clears the crossbar.
The ball's initial horizontal speed is 26*cos(30.6) m/s and there is no horizontal acceleration.
The ball's initial vertical speed is 26*sin(30.6) m/s and there is a vertical acceleration of -g
Resolving horizontally to find the time it takes to reach the crossbar:
s = ut + a(t^2)/2
52 = 26*cos(30.6)*t
t = 52/(26*cos(30.6)) = 2.324 seconds
Resolving vertically to calculate the height of the ball at the time it is passing the crossbar:
s = ut + a(t^2)/2
s = 26*sin(30.6)*t - (g/2)*(t^2)
s = 4.271 Meters
So the ball clears the post at a height of 1.221 Meters.