3. It’s fourth down in the big football game, and the game is tied with 2 seconds to go! You are called on to win the game with the biggest field goal of your life, from 36 meters. The crowd holds its collective breath as the ball is spotted, and you kick it. It leaves your foot with a speed of 22 m/s at an angle of 52º above the ground. The crossbar on the goal post is 3.05 meters high.
A. How long does the ball take to reach the goal post? (5 points)
B. What is the ball’s speed when it reaches the goal post? (5 points)
C. By how much is the ball above or below the crossbar when it reaches the plane of the goal post? (5 points)
in the x-direction,
vx = 22*cos52º = 13.5m/s
it travels 36m in 36/13.54 = 2.66 seconds
neglecting air resistance
vy = 22*sin52º - 9.8t = 17.34 - 9.8*2.66 = -8.73
vx is still 13.54
use pythagorean theorem to add the vectors
y = 0 + 17.34t - 4.9t^2 = 17.34*2.66 - 4.9*2.66^2 = 11.45
so, it clears the bar
To solve this problem, we can use the equations of motion for projectile motion. Let's break down the steps for each part of the question:
A. How long does the ball take to reach the goal post?
To find the time it takes for the ball to reach the goal post, we can use the equation for the vertical motion of a projectile:
v_y = v_i * sin(theta) - g * t
Here, v_y is the vertical component of the ball's velocity, v_i is the initial velocity, theta is the launch angle, g is the acceleration due to gravity, and t is the time. We know the initial velocity of the ball (22 m/s), the launch angle (52º), and the acceleration due to gravity (9.8 m/s^2).
First, let's find v_y:
v_y = v_i * sin(theta)
v_y = 22 m/s * sin(52º)
v_y ≈ 16.86 m/s (rounded to two decimal places)
Now, we can solve the equation for t:
v_y = v_i * sin(theta) - g * t
16.86 m/s = 22 m/s * sin(52º) - 9.8 m/s^2 * t
We can rearrange this equation to solve for t:
t = (v_i * sin(theta) - v_y) / g
t = (22 m/s * sin(52º) - 16.86 m/s) / 9.8 m/s^2
Using a calculator, we find:
t ≈ 1.18 seconds (rounded to two decimal places)
Therefore, the ball takes approximately 1.18 seconds to reach the goal post.
B. What is the ball’s speed when it reaches the goal post?
To find the speed of the ball when it reaches the goal post, we can use the equation for the horizontal motion of a projectile:
v_x = v_i * cos(theta)
Here, v_x is the horizontal component of the ball's velocity, v_i is the initial velocity, and theta is the launch angle. We know the initial velocity of the ball (22 m/s) and the launch angle (52º).
Let's calculate v_x:
v_x = v_i * cos(theta)
v_x = 22 m/s * cos(52º)
v_x ≈ 13.34 m/s (rounded to two decimal places)
Therefore, the ball's speed when it reaches the goal post is approximately 13.34 m/s.
C. By how much is the ball above or below the crossbar when it reaches the plane of the goal post?
To find how much the ball is above or below the crossbar, we need to determine the vertical position of the ball at the time it reaches the goal post.
Using the equation for vertical motion:
y = y_0 + v_i * sin(theta) * t - (1/2) * g * t^2
Here, y is the vertical position of the ball, y_0 is the initial vertical position (0 in this case as the ball is kicked from the ground), v_i is the initial velocity, theta is the launch angle, g is the acceleration due to gravity, and t is the time. We know the initial velocity of the ball (22 m/s), the launch angle (52º), the acceleration due to gravity (9.8 m/s^2), and the time it takes to reach the goal post (1.18 seconds).
Substituting these values into the equation, we can calculate y:
y = 0 + 22 m/s * sin(52º) * 1.18 s - (1/2) * 9.8 m/s^2 * (1.18 s)^2
Using a calculator, we find:
y ≈ 9.69 meters (rounded to two decimal places)
Since the crossbar is 3.05 meters high, we can determine the difference between the ball's position and the height of the crossbar:
Difference = y - crossbar height
Difference = 9.69 m - 3.05 m
Difference ≈ 6.64 meters (rounded to two decimal places)
Therefore, the ball is approximately 6.64 meters above the crossbar when it reaches the plane of the goal post.