What is the maximum height of the stadium at its back wall?

In summary, Ted Williams hits a baseball with an initial velocity of 120 miles per hour at an angle of 35 degrees to the horizontal. He claims that if there was no air resistance, he could have hit the ball out of the stadium. Assuming this is true, the maximum height of the stadium's back wall 565 feet from home plate would need to be 151.35 feet for the ball to just pass over it. The calculations were done using the equations for horizontal and vertical motion, with a horizontal acceleration of 0 and a vertical acceleration of -32.2 ft/s^2. The resulting values were 3.919 seconds for the time and 151.35 feet for the horizontal distance.
  • #1
maiad
102
0

Homework Statement


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?

You may need:
9.8 m/s2 = 32.2 ft/s2
1 mile = 5280 ft

Homework Equations


Xf=Xi+Vit+0.5a(t^2)

The Attempt at a Solution


Horizontal:
a=0
Vi=176cos35
xi=0
Xf=565

Vertical:
a=-32.2ft/s^2
Vi=176sin35
Xi=3 ft

i simply plugged the horizontal values in and solved for t which was 3.918

i then used t and the vertical components and simply plugged it in and gave me Xf being 151.34ft ft which is wrong... This is a online grading so they might be picky about the accuracu but i just want to be sure I'm doing the steps right.
 
Last edited:
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  • #2
The steps seem OK. The values are off in the last digit: I get 3.919 and 151.35.
 
  • #3
Why are you bumping this?
 
  • #4
if voko's post didn't help, then you should explain why :)
 
  • #5
woops wrong post i bumped LOL. But yea... still giving me me as the wrong answer for some reason... i think i'll just ask my proff if there's something wrong with the grading system
 

1. What is the difference between 2D and 3D kinematics?

2D kinematics refers to the study of motion in two dimensions, typically represented on a flat surface or plane. This includes motion along the x and y axes. On the other hand, 3D kinematics involves the study of motion in three dimensions, which includes motion along the x, y, and z axes.

2. How is displacement calculated in 2D kinematics?

In 2D kinematics, displacement is calculated using the Pythagorean theorem. This means that the magnitude of the displacement is equal to the square root of the sum of the squares of the change in position along the x and y axes. The direction of the displacement can be determined using trigonometric functions.

3. What is the difference between speed and velocity in 2D kinematics?

Speed is a scalar quantity that refers to the rate at which an object moves, while velocity is a vector quantity that includes both the speed and direction of an object's motion. In 2D kinematics, speed can be calculated by dividing the total distance travelled by the total time taken, while velocity can be calculated by dividing the total displacement by the total time taken.

4. How do you calculate acceleration in 2D kinematics?

Acceleration in 2D kinematics is calculated using the same formula as in 1D kinematics, which is a = (vf - vi)/t, where vf and vi are the final and initial velocities, respectively, and t is the time interval. However, in 2D kinematics, acceleration can also be calculated as the rate of change of velocity along the x and y axes.

5. What are some common equations used in 2D kinematics?

Some common equations used in 2D kinematics include the equations for displacement, velocity, and acceleration, as well as the equations for projectile motion, such as the range equation (R = (v^2 * sin(2θ))/g) and the maximum height equation (H = (v^2 * sin^2(θ))/(2g)). Additionally, the equations for circular motion, such as the centripetal acceleration equation (a = v^2/r) and the centripetal force equation (F = mv^2/r), are also commonly used in 2D kinematics.

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