Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of è = 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!'

Question

Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of è = 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? You may need:

•9.8 m/s2 = 32.2 ft/s2
•1 mile = 5280 ft

h =................?

Answer 1

Vo = 176 Ft/s @ 35 deg.

Xo = hor. = 176cos35 = 144.17 Ft/s.
Yo = ver. = 176sin35 = 100.95 Ft/s.

h = 3 + (Vf^2 - Yo^2) / 2g,
h = 3 +(0 - (100.95)^2) / -64.4 =
161.2 Ft. above ground.

Answer 2

To find the maximum height of the stadium's back wall, we need to analyze the projectile motion of the baseball. Since air resistance is negligible, we can assume the only force acting on the ball is gravity.

We can break down the initial velocity of the ball into its horizontal and vertical components. Firstly, let's find the initial vertical velocity (v0y) and initial horizontal velocity (v0x) using the given angle and initial velocity.

v0x = v0 * cos(θ)
v0x = 176 ft/s * cos(35°)
v0x = 144.39 ft/s (approx)

v0y = v0 * sin(θ)
v0y = 176 ft/s * sin(35°)
v0y = 100.05 ft/s (approx)

We know that the time taken for the ball to travel from the home plate to the stadium's back wall (565 ft away) can be found using the horizontal component of velocity:

Time = Distance / Velocity
Time = 565 ft / 144.39 ft/s
Time = 3.91 seconds (approx)

Next, we can find the time it takes for the ball to reach its maximum height, considering the symmetry of projectile motion. The time to reach the maximum height would be half of the total flight time:

Time to reach maximum height = 3.91 seconds / 2
Time to reach maximum height = 1.96 seconds (approx)

Using the equation for vertical displacement, we can find the maximum height (h) the ball reaches:

h = v0y * t - (0.5 * g * t^2)

where g is the acceleration due to gravity (32.2 ft/s^2).

h = 100.05 ft/s * 1.96 s - (0.5 * 32.2 ft/s^2 * (1.96 s)^2)
h = 196.1 ft - (0.5 * 32.2 ft/s^2 * 3.84 s^2)
h = 196.1 ft - 62.48 ft
h = 133.62 ft (approx)

Therefore, the maximum height of the stadium's back wall would be approximately 133.62 feet for the ball to just pass over it if there was no air resistance.