How Do You Calculate the Minimum Speed of a Baseball for a Home Run in Physics?

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Homework Help Overview

The problem involves calculating the minimum speed of a baseball for it to clear a 7.0m-high fence located 120m from home plate, given that the ball is hit from a height of 1.0m at a 30-degree angle. Participants are exploring the implications of the initial height and the trajectory of the ball.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations, questioning how to relate the time of flight for both vertical and horizontal components. Some suggest breaking the problem into smaller parts and analyzing the trajectory at maximum height. Others express uncertainty about how to find the initial velocities without additional information.

Discussion Status

There is ongoing exploration of different approaches to the problem, with some participants offering hints and guidance on how to set up equations based on the ball's trajectory. Multiple interpretations of the problem are being discussed, particularly regarding the use of initial conditions and the relationship between vertical and horizontal motion.

Contextual Notes

Participants mention constraints such as the initial height not being zero and the need to avoid using the range equation due to classroom guidelines. There are also concerns about the implications of struggling with this problem on future academic aspirations.

Ritzycat
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I'm struggling to do this problem.

"When Babe Ruth hit a homer over the 7.0m -high right-field fence 120m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0m above the ground and its path initially made a 30∘ angle with the ground."

My work:
Since initial Y is not 0, the range equation cannot be used.

initial Y: 1m
final Y: 7m
initial Y velocity: ?
final Y velocity: ?
a = -9.8m/s
t = ? (but is same as T for X)

initial X: 0m
final X: 120m
initial X velocity: ?
final Y velocity: ?
a = 0
t = ? (but is same as T for Y)

I have attempted manipulating the 3 main kinematic equations to try to set T equal to each other for each component but I am empty handed.

Can someone help me step on the right foot?

Also, if I am struggling with this problem, is it still possible for me to obtain a PhD in physics one day?
 
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Ritzycat said:
I'm struggling to do this problem.

"When Babe Ruth hit a homer over the 7.0m -high right-field fence 120m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0m above the ground and its path initially made a 30∘ angle with the ground."

My work:
Since initial Y is not 0, the range equation cannot be used.

initial Y: 1m
final Y: 7m
initial Y velocity: ?
final Y velocity: ?
a = -9.8m/s
t = ? (but is same as T for X)

initial X: 0m
final X: 120m
initial X velocity: ?
final Y velocity: ?
a = 0
t = ? (but is same as T for Y)

I have attempted manipulating the 3 main kinematic equations to try to set T equal to each other for each component but I am empty handed.

Can someone help me step on the right foot?

Instead of manipulating formulas, why not put some thought into solving this problem?

Hint: it may take more than one step to work out the solution. Break up the whole problem into smaller sub-problems, and solve each in turn.

Try analyzing the problem starting at the moment the ball leaves the bat. You know the initial height of the ball and the angle at which it leaves the bat.
Take the initial speed as V. Write an equation which tells you how long it takes the ball to reach its maximum height, using this speed V.
How far does the ball travel horizontally from the bat to the point of maximum altitude?

Once you get a handle on the first part, see if you can set up equations for the trajectory of the ball as it falls back to earth.

Also, if I am struggling with this problem, is it still possible for me to obtain a PhD in physics one day?

You're a long way away from a Physics PhD. You need to concentrate on Intro Physics for the moment.
 
Here's an equation for Y when it reaches maximum height (where y-velocity equals 0m/s)

Initial Y velocity = V sin 30
Initial X velocity = V cos 30

0m/s = (V)(sin 30) - (9.8m/s^2)(t)

Then I can find X distance
x = (V)(cos 30)(t)

Where T here equals T in the first equation, whatever that may be.

I have no idea how to go about finding the overall initial velocity, initial X velocity, or initial Y velocity, without having one of those three in addition to the 30 degree angle, so I could use trig to find the missing parts.
 
Last edited:
Ritzycat said:
Here's an equation for Y when it reaches maximum height (where y-velocity equals 0m/s)

Initial Y velocity = V sin 30
Initial X velocity = V cos 30

0m/s = (V)(sin 30) - (9.8m/s^2)(t)

Then I can find X distance
x = (V)(cos 30)(t)

Where T here equals T in the first equation, whatever that may be.

I have no idea how to go about finding the overall initial velocity, initial X velocity, or initial Y velocity, without having one of those three in addition to the 30 degree angle, so I could use trig to find the missing parts.

The solution you have to the first part looks good.

Even though the ball has reached the top of its trajectory, it is still traveling horizontally, and it must clear the fence as described in the OP. Can you write some equations from this point in the ball's travel such that if the ball clears the top of that 7 meter fence 120m from home plate, you can find the minimum initial velocity the ball had on leaving the bat? Give it a shot.
 
7m = yo + 1/2(-9.8 m/s^2)(t^2)
120m = xo + volt

from the maximum height to the end
 
Knowing that x0 and y0 are the location of the maximum altitude of the ball's trajectory, you should also establish what x0 and y0 are relative to the point where the ball was hit. When you do this, you should be able to solve for the initial velocity of the ball when it is batted, in order for it to clear the top of the fence.
 
If I'm understanding what you said correctly, I will set Y(initial) of the second part (at the ball's maximum height) equal to Y(final) of the first part of the motion.

7m = Yo + 1/2(-9.8 m/s^2)(t^2)

7m + (4.9m/s^2)(t^2) = Yf

7m + (4.9m/s^2)(t^2) = 1m + (v)(sin 30)(t) - (4.9m/s^2)(t^2)

6m + (9.8m/s^2)(t^2) = (v)(sin 30)(t)

6m + (9.8m/s^2)(t) / (sin 30) = V
 
I took a different approach and manipulated a few steps from when we derivated the Range Equation in class. My answer was close enough to get full credit (although it was not the most ideal or correct way.) My teacher showed me what I was confused with and the "proper" method to do it, since she says the Range Equation is looked down upon and if we use it on our tests or on the AP test we get no credit for the question. Seems a bit silly to me.

Thanks for your help, you got me on the right track, I think I understand how to approach these types of problems.
 

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