Solving a Baseball Word Problem: Time in Air & Height of Ball's Path

  • Thread starter RHSLilSweetie07
  • Start date
  • Tags
    Word problem
In summary: I had a better understanding of what was going on and could more easily manipulate the formulas. So I'm confident that I could have found the correct answer if I needed to. In summary, the ball is caught 65.0 m from the thrower. The acceleration of gravity is 9.81 m/s^2.
  • #1
RHSLilSweetie07
2
0
A baseball is thrown at an angle of 24 degrees relative to the ground at a speed of 29.3 m/s. The ball is caught 65.0 m from the thrower. The acceleration of gravity is 9.81 m/s^2.

a) How long is it in the air? Answer in units of s.

b) How high is the tallest spot in the ball's path? Answer in units of m.

I have no clue on how to solve this problem. Will someone please help, or at least give me the formula to use.

Thanks
 
Physics news on Phys.org
  • #2
When I help people I know with ballistics-esque problems I always try to get them to understand that the ball doesn't change velocity in the x direction , because we ignore air friction for simpler physics...and since the ball has gravity applied to it, it will be what 'limits' the range. Like.. when the scenario is taking place, the ball is moving, what dictates when it stops is when the y component reaches its desired height, now since it doesn't mention anything about the ball being caught by someone taller or shorter.. or landing on the ground, you have to assume that the level from which the ball is released is the same as where it is caught, AND that this is the 0 for your height.

First, I'd find the vector components of the velocity of the ball. Since velocity is a vector, and you're taking it at the instant that the ball is released, you do not need to fret over the arc path that the ball will travel.
So, you can treat it like finding the x/y components of any simple vector.
v_x = 29.3 cos(24)
v_y = 29.3 sin(24)

Now, you want to figure out the amount of time it takes for the ball to accelerate from its v_y to 0 , since the same acceleration is being applied and the displacement of the ball is the same, then the time up will equal the time down, in one instance you're subtracting your v_y and in the other you're subtracting 0 from v_y , the one where you subtract v_y will have a negative accel, the other a positive.. so in the end it cancels out.

Now, since we assume the v_x doesn't change then the 'range' of the object is going to be the displacement that occurs over the time you find for the object to go up and down, at a rate of v_x.

Sorry if the wording is a little tricky, I'd rather not just hand you an answer.

using some of the simple formulas you can work to your answer. One place you might want to go is http://scienceworld.wolfram.com/physics/Ballistics.html

that page is pretty much blank, but the 'related to ' links will help you.. I didn't want to just put a bunch of them in here.

If this doesn't give you the nudge you need then just ask for clarification or something, or let me know how far you got and i'll see if i can't nudge a little more..
 
  • #3
If your physics class is calculus based, pls see the following link for a tutorial I wrote specifically for this forum that illustrates a simple, structured methodology for dealing with problems like this. It includes a sample projectile problem (3rd attachment, I think), which is what you have.

https://www.physicsforums.com/showthread.php?t=93670
 
  • #4
Last edited:
  • #5
I think perhaps the more important question is whether you understand the principles involved and the methodology to achieve the anwser.:smile:
 
  • #6
hotvette: I doubt the class is directly calculus based, its probably a HS level or so class, but that is a nice one, I'm going to use it :-D

RHS : Hotvette is exactly right, if all you know is whether or not your numbers are right, then all you know is the answers for this problem. If you ensure that you understand how to manipulate formulas and stuff to get to your answer, then you can answer any problem (on the topic).

I got a different number for the time. If you know acceleration, and you know the change in velocity then what you want to do is find a formula that uses a (accel), (delta) v (change in veloc), and delta t (change in time).

One thing I did in my first physics class is I would always crack open my book at least once a day and write the equations in the 'base forms' that they were given (not like there is a 'base form' but the way they were given) and i would just algebra it up and solve for each variable in each one.. and then use substitution to solve for variables across equations. After a few days I'd know the equations in the book, and how to use them together to get to where I wanted to be (<variable desired> = <expression with variables that have known values>)

you might want to try and find your own method of solving it rather than going to that place and just copying their work. If you do that, then all you did was follow steps, but you're not given steps when the problems matter a little more.
 
  • #7
stmoe said:
One thing I did in my first physics class is I would always crack open my book at least once a day and write the equations in...

As a study aid for final exams, I used to write tutorials with detailed example problems explaining each step!
 

What is the formula for calculating the time in air of a baseball?

The formula for calculating the time in air of a baseball is t = (2*v*sinθ)/g, where t is the time in seconds, v is the initial velocity of the ball in meters per second, θ is the angle of launch in degrees, and g is the acceleration due to gravity (9.8 m/s²).

How do you calculate the height of a baseball's path?

The formula for calculating the height of a baseball's path is h = (v²*sin²θ)/(2*g), where h is the height in meters, v is the initial velocity of the ball in meters per second, θ is the angle of launch in degrees, and g is the acceleration due to gravity (9.8 m/s²).

What units are used for time and height in air calculations?

The units used for time in air calculations are seconds (s) and the units for height in air calculations are meters (m).

How do you convert miles per hour (mph) to meters per second (m/s)?

To convert miles per hour (mph) to meters per second (m/s), multiply the mph value by 0.44704. For example, 60 mph is equivalent to 26.8224 m/s.

What factors can affect the accuracy of calculations for time in air and height of a baseball's path?

The accuracy of calculations for time in air and height of a baseball's path can be affected by factors such as air resistance, wind speed and direction, and the condition of the baseball (e.g. weight, shape, etc.). Additionally, human error in measuring the initial velocity and angle of launch can also impact the accuracy of the calculations.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
977
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
764
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top