Joey kicks a football with an initial velocity of 42'/sec at an angle of 35 degrees to the horizon. what is the longest field goal he can make if the crossbar is 10' high?
at time t,
x = 42 cos35 t
y = 42 sin35 t - 16t^2 = 24t - 16t^2
y = 10 when
24t - 16t^2 = 10
16t^2 - 24t + 10 = 0
Houston, we have a problem. Are those numbers right?
To determine the longest field goal Joey can make, we will find the horizontal distance the football travels before reaching the crossbar.
Let's first break down the initial velocity into its horizontal and vertical components.
Horizontal Component:
v₀x = v₀ * cosθ
v₀x = 42 ft/sec * cos(35°)
v₀x = 42 ft/sec * 0.819
v₀x ≈ 34.38 ft/sec
Vertical Component:
v₀y = v₀ * sinθ
v₀y = 42 ft/sec * sin(35°)
v₀y = 42 ft/sec * 0.574
v₀y ≈ 24.12 ft/sec
Now, we need to calculate the time it takes for the ball to reach its highest point. At this point, the vertical velocity becomes zero before it starts to descend.
Using the vertical kinematic equation:
v₂y = v₀y - g * t
0 = 24.12 ft/sec - 32.17 ft/sec² * t
Solving for t:
t = v₀y / g
t = 24.12 ft/sec / 32.17 ft/sec²
t ≈ 0.75 sec
Now we can calculate the total time it takes for the ball to reach the crossbar by doubling the time it takes to reach the highest point.
Total time = 2 * t
Total time ≈ 2 * 0.75 sec
Total time ≈ 1.5 sec
Finally, we can calculate the horizontal distance the football travels by multiplying the horizontal component of velocity by the total time of flight.
Horizontal distance = v₀x * t
Horizontal distance ≈ 34.38 ft/sec * 1.5 sec
Horizontal distance ≈ 51.57 ft
Therefore, Joey can make the longest field goal up to a horizontal distance of approximately 51.57 feet.
To find the longest field goal Joey can make, we need to determine the horizontal distance the football travels before hitting the ground.
First, we can decompose the initial velocity into its horizontal and vertical components. The horizontal component is given by Vx = V * cos(theta), where V is the initial velocity (42'/sec) and theta is the angle (35 degrees). Substitute the values we know:
Vx = 42' * cos(35 degrees)
Next, we need to determine the time it takes for the football to hit the ground. To do this, we'll use the vertical component of the initial velocity. The vertical component is given by Vy = V * sin(theta), where V is the initial velocity (42'/sec) and theta is the angle (35 degrees). Substitute the values we know:
Vy = 42' * sin(35 degrees)
Considering the motion in the vertical direction, the time it takes to reach the highest point of the trajectory is given by t = Vy / g, where g is the acceleration due to gravity (32.2 ft/sec^2). Substitute the values we know:
t = (42' * sin(35 degrees)) / 32.2 ft/sec^2
After finding the time it takes to reach the highest point, we can determine the total time of flight by doubling it:
T = 2 * [(42' * sin(35 degrees)) / 32.2 ft/sec^2]
Now, we can find the horizontal distance traveled (x) during this total time by multiplying the horizontal component of velocity (Vx) by the total time of flight (T):
x = Vx * T
The longest field goal Joey can make is the horizontal distance x where the football hits the ground. However, there is an additional constraint that the football must clear the 10' high crossbar. So we'll need to find the maximum height (H) the ball reaches and make sure it's higher than 10'.
To find the maximum height (H), we can use the equation H = (Vy^2) / (2g), where Vy is the vertical component of velocity and g is the acceleration due to gravity:
H = [(42' * sin(35 degrees))^2] / (2 * 32.2 ft/sec^2)
Finally, we can calculate the longest field goal distance (x) by considering the horizontal distance traveled during the total time of flight, but making sure the height (H) is greater than 10':
x = Vx * T, subject to H > 10'
Now you can plug in the values and calculate the longest field goal Joey can make.