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?

The ball's horizontal speed is 176 cos35° = 144 ft/s

So, it would take 565/144 = 3.92 seconds to travel to the wall.

The height of the ball is
y = 3 + 176 sin35° t - 16t^2
= 3 + 101t - 16t^2
So, what is y when t = 3.92?

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 concept of projectile motion.

First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component Vx remains constant throughout the motion, while the vertical component Vy changes due to the effect of gravity.

Given that the initial velocity V = 176 ft/s is at an angle θ = 35 degrees to the horizontal, we can calculate the horizontal and vertical components as follows:

Vx = V * cos(θ)
Vy = V * sin(θ)

Vx = 176 * cos(35°) ≈ 144.63 ft/s
Vy = 176 * sin(35°) ≈ 98.22 ft/s

Now, let's analyze the vertical motion of the baseball. The time of flight can be determined using the formula:

t = (2 * Vy) / g

where g is the acceleration due to gravity (approximately 32.2 ft/s²). Substituting the values:

t = (2 * 98.22) / 32.2 ≈ 6.08 seconds

The maximum height of the ball can be found by using the equation:

y = yo + Vy * t - (1/2) * g * t²

where yo is the initial height and y is the maximum height. In this case, yo = 3 ft and y will be the maximum height.

y = 3 + 98.22 * 6.08 - (1/2) * 32.2 * (6.08)²

y ≈ 304.03 ft

So, if there was no air resistance, the maximum height of the stadium at its back wall (x = 565 ft) such that the ball would just pass over it would be approximately 304.03 feet.