Baseball: finding trajectory of ball

[Naeem]

Q.Ted Williams hits a baseball with an initial velocity of 120 miles per hour (176 ft/s) at an angle of q = 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 = feet

Can somebody tell me what have to find step by step in this problem.
The help provided with this problem isn't very helpful..

Thanks

 

[so-crates]

x(t) = ?
y(t) = ?

x(t1) = 565, t1 = ?
y(t1) = ?

 

[thepatientmental]

for your question! Let's break down the steps needed to solve this problem:

Step 1: Understand the given information
We are given the initial velocity of the ball (120 miles per hour or 176 ft/s), the angle at which it was hit (35 degrees), and the distance it traveled (420 feet). We are also told that the ball was struck 3 feet above home plate.

Step 2: Convert units
Since the given velocity is in miles per hour and the distance is in feet, we need to convert the velocity to feet per second. We can do this by multiplying 120 miles per hour by 5280 feet per mile and dividing by 3600 seconds per hour. This gives us 176 ft/s, which is the same as the given velocity.

Step 3: Find the time of flight
Using the given initial velocity and angle, we can use the kinematic equations to find the time of flight of the ball. The equation we will use is t = (2v*sin(q))/g, where v is the initial velocity, q is the angle, and g is the acceleration due to gravity (32.2 ft/s^2). Plugging in the given values, we get t = (2*176*sin(35))/32.2 = 3.35 seconds.

Step 4: Find the maximum height
To find the maximum height, we can use the equation h = v^2*sin^2(q)/(2g), where v is the initial velocity, q is the angle, and g is the acceleration due to gravity. Plugging in the given values, we get h = (176^2*sin^2(35))/(2*32.2) = 89.19 feet. This means that the maximum height of the ball's trajectory is 89.19 feet.

Step 5: Find the distance traveled by the ball at the maximum height
To find the distance traveled by the ball at the maximum height, we can use the equation d = v^2*sin(2q)/g, where v is the initial velocity, q is the angle, and g is the acceleration due to gravity. Plugging in the given values, we get d = (176^2*sin(2*35))/32.2 = 563.56 feet. This means that the ball traveled 563.56 feet horizontally at the maximum height.

Step 6: Find the height of the stadium