How Far Does a Ball Fall Vertically in Projectile Motion?

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a ball thrown horizontally and its vertical displacement by the time it reaches home plate, which is 60 feet away. The acceleration due to gravity is given as 32 ft/s², and the initial horizontal velocity is provided as 96.9 mi/h, converted to 142.12 ft/s.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to determine the time of flight and vertical displacement. There is uncertainty about which equations to apply and the initial vertical velocity.

Discussion Status

Some participants have offered guidance on identifying the correct initial vertical velocity and the direction of acceleration. There is a mix of attempts to clarify the problem and check assumptions, with one participant expressing confusion about the overall approach.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an ongoing exploration of the implications of horizontal launch conditions on vertical motion.

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Projectile Motion and Level Horizontal Range.

Homework Statement


The acceleration of gravity is 32ft/s ^2.
If a pitch were thrown horizontally with a velocity of 96.9 mi/h, how far would the ball fall vertically by the time it reached home plate, 60ft away. Answer in units of ft.
So speed is 142.12 ft/s.

Homework Equations


Vx = Vxo + Axt
Y= Yo + VyoT + 1/2AyT^2 or Vy^2 = Vyo^2 + 2Ay(Y-Yo)

For the second equation, don't know which one to use.


The Attempt at a Solution


I got the time it took to reach 60 feet by using the first equation. T= 4.441. I don't know where to go from there. I think I use this equation: Y= Yo + VyoT + 1/2AyT^2

but I'm not sure. Any help would be appreciated greatly.

Edit: actually t= .422 seconds.
 
Last edited:
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Anybody?
 
sorry
 
Last edited:
I'm not sure what you are asking. Sorry to be such a pain.
 
Sorry, ok so you know how long it takes for the ball to get from release to home. It travels a distance in the -y direction in that time. Thats what your looking for.
 
using this equation: Y= Yo + VyoT + 1/2AyT^2, I got 57.151. That definitely is not right or is it?
 
probably not what are you using for Vy at t=0? and if y=0 at the pitchers release what direction is acceleration in the y direction?
 
I cannot figure out this problem. Maybe I got the time wrong. I don't even know what I'm looking for. Can somebody please clarify some stuff for me. I just want to know how to do it.
 
harvellt said:
probably not what are you using for Vy at t=0?

You mean Vyo, I used the original speed the ball was thrown, 142.12 feet/s. Is that wrong?
 
  • #10
yes that is wrong. That is the Vox, what is the Voy if the ball is thrown horizontally?
 
  • #11
zero?
 
  • #12
yep, and then what is the acceleration of the ball?
 
  • #13
OMG I just got it. Voy is 0! So it is -y = 1/2 (32) (.422)^2. Which gives me 2.849. Which is right in the computer. Thank you for your help!
 
  • #14
no prob!
 

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