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 =................?
Check your 9-17-11,1:12pm post for
solution.
thats wrong.
To determine the maximum height of the stadium at its back wall, we need to analyze the projectile motion of the ball. We can break down the problem into horizontal and vertical components.
Let's begin with the horizontal component. We know that the ball travels a horizontal distance of 565 feet from home plate to the back wall of the stadium. Since there is no air resistance, the horizontal velocity remains constant throughout the entire flight. Thus, we can use the formula:
distance = velocity * time
The horizontal velocity (Vx) can be calculated using the initial velocity (Vi) and the launch angle (θ):
Vx = Vi * cos(θ)
In this case, Vi is given as 176 ft/s and the launch angle (θ) is given as 35 degrees. Plugging the values into the formula, we get:
Vx = 176 ft/s * cos(35 degrees)
Next, we can determine the time it takes for the ball to travel the horizontal distance (565 feet). Since the horizontal velocity is constant, we can use the equation:
time = distance / velocity
Plugging in the values, we get:
time = 565 ft / Vx
Now, let's move on to the vertical component. We want to find the maximum height (h) at which the ball passes over the back wall of the stadium. At the maximum height, the vertical velocity (Vy) becomes zero. We can use the equation:
Vy = Vi * sin(θ) - (g * t)
where g is the acceleration due to gravity (32.2 ft/s^2) and t is the time it takes for the ball to reach the maximum height.
Setting Vy as zero and solving for t, we get:
t = (Vi * sin(θ)) / g
Plugging in the values, we get:
t = (176 ft/s * sin(35 degrees)) / 32.2 ft/s^2
Finally, we can calculate the maximum height using the equation:
h = Vi * sin(θ) * t - (1/2) * g * t^2
Plugging in the values, we get:
h = (176 ft/s * sin(35 degrees) * t) - (0.5 * 32.2 ft/s^2 * t^2)
Thus, by solving for h, we can determine the maximum height of the stadium at its back wall such that the ball would just pass over it.