Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of θ = 35 degrees to the horizontal. The ball is struck 3 feet above home plate. You watch as the ball goes over the outfield wall 420 feet away and lands in the bleachers. After you congratulate Ted on his hit he tells you, 'You think that was something, if there was no air resistance I could have hit that ball clear out of the stadium!'

Assuming Ted is correct, what is the maximum height of the stadium at its back wall x = 565 feet from home plate, such that the ball would just pass over it?

equations:

y= 0 + v*sin(35)*t - 1/2*g*t^2

x= v*cos(35)*t
[we have x=565, v=176 ft/s, so t=3.918964]

plug t into y equation, we get y [use gravity in ft/s^2]

and then add 3ft b/c the ball was hit from 3ft above the ground,

so the height = y+3 = 148.35 + 3 = 151.35 ft

To find the maximum height of the stadium at the back wall such that the ball would just pass over it, we can use the projectile motion equations.

1. Convert the initial velocity from miles per hour to feet per second.
Given: Initial velocity (v) = 120 miles per hour = 176 ft/s

2. Calculate the time of flight (t):
The time of flight is the total time taken by the ball to reach the back wall. We can use the horizontal distance traveled (x) and the horizontal component of velocity (v_x) to find the time.
Given: Horizontal distance (x) = 565 feet
Horizontal component of velocity (v_x) = v * cos(θ)
where θ = 35 degrees
Time of flight (t) = x / v_x

3. Calculate the vertical component of velocity (v_y):
The vertical component of velocity is given by v_y = v * sin(θ).

4. Calculate the time taken to reach the maximum height (t_max):
To reach the maximum height, the vertical component of velocity becomes 0.
Therefore, t_max = v_y / g
where g is the acceleration due to gravity (32.2 ft/s^2).

5. Calculate the maximum height (h_max):
To find the maximum height, we can use the formula:
h_max = v_y * t_max - (1/2) * g * t_max^2

Now let's calculate the maximum height of the stadium:

Step 1: Convert the initial velocity to feet per second:
v = 120 miles per hour = 176 ft/s

Step 2: Calculate the time of flight (t):
v_x = v * cos(θ) = 176 ft/s * cos(35°)
≈ 144.028 ft/s

t = x / v_x = 565 ft / 144.028 ft/s
≈ 3.925 s

Step 3: Calculate the vertical component of velocity (v_y):
v_y = v * sin(θ) = 176 ft/s * sin(35°)
≈ 100.6 ft/s

Step 4: Calculate the time taken to reach the maximum height (t_max):
t_max = v_y / g = 100.6 ft/s / 32.2 ft/s^2
≈ 3.124 s

Step 5: Calculate the maximum height (h_max):
h_max = v_y * t_max - (1/2) * g * t_max^2
= 100.6 ft/s * 3.124 s - (1/2) * 32.2 ft/s^2 * (3.124 s)^2
≈ 156.643 ft

Therefore, the maximum height of the stadium at its back wall (565 feet from home plate) such that the ball would just pass over it is approximately 156.643 feet.

To determine the maximum height of the stadium at its back wall, we need to analyze the projectile motion of the baseball and find the height at that distance where the ball would just pass over it.

Here's how we can approach this problem:

1. Break down the initial velocity of the baseball into its horizontal and vertical components. The vertical component can be calculated as V_y = V * sin(θ), where V is the initial velocity of the ball, and θ is the angle at which it was hit. Therefore, V_y = 176 ft/s * sin(35°).

2. Calculate the time taken by the ball to reach a horizontal distance of 420 feet. We can use the horizontal component of velocity (V_x = V * cos(θ)) and the given distance to calculate the time (t) as t = distance / V_x.

3. Since we are assuming no air resistance, we can use the equation of motion to find the maximum height (H_max) reached by the ball at time t. The equation is H = V_y * t - (1/2) * g * t^2, where g is the acceleration due to gravity (32.17 ft/s^2). Plug in the values to calculate H_max.

4. Now, we need to determine how high the back wall of the stadium would be (x = 565 feet) so that the ball just passes over it when it lands. We can start by assuming a height (H_wall) and calculate the horizontal distance covered by the ball (D = V_x * t) at that height.

5. If D is less than x, we need to increase the height. If D is greater than x, we need to decrease the height. We can repeat the process, adjusting the height until D is very close to x.

By following these steps, we can determine the maximum height of the stadium at its back wall such that the ball would just pass over it.

the horizontal velocity of the ball is

___ Vh = 176 cos(35º)

the vertical velocity of the ball is
___ Vv = 176 sin(35º)

it takes the ball 565 / Vh
seconds to reach the back wall

the time to the peak of the ball's
trajectory is ___ Vv / g

the difference between the peak time and the flight time to the wall is the time the ball has to fall from the peak height
___ this will give you the height at the wall