Projectile Motion of a Baseball Problem

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a baseball hit from ground level. The ball's trajectory includes reaching a maximum height after 3.5 seconds and clearing a fence located 95.8 meters away after an additional 3 seconds. Participants are tasked with determining the maximum height reached by the ball, the height of the fence, and the distance the ball travels beyond the fence before hitting the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to find the maximum height and the height of the fence. Some express confusion about their calculations and the relationships between the given times and distances.

Discussion Status

Some participants have provided hints and guidance on how to approach the problem, particularly regarding the significance of the time at maximum height and the symmetry of the projectile's motion. However, there is no explicit consensus on the correct approach or final answers, as participants continue to explore different interpretations and calculations.

Contextual Notes

Participants note that the original poster's calculations seem inconsistent, particularly regarding the total time of flight and the relationship between the distances involved. There is an emphasis on the need for detailed solutions to facilitate further assistance.

mcg5132
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1.
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.5 s after being hit. Then 3.0 s after reaching its maximum height, the ball barely clears a fence that is 95.8 m from where it was hit. Assume the ground is level. What maximum height above ground level is reached by the ball? How high is the fence? How far beyond the fence does the ball strike the ground?





2. X-Xo=(Vocosx0)t
Y-Yo=V0yt-1/2gt^2
V^2y=(Vosinxo)^2-2g(Y-Yo)





3.
 
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where is your attempt to solve the question?
 
No idea where to even begin. None of my numbers are making any sense when I use these equations.
 
First, have a look at the figure I made:

http://img6.imageshack.us/img6/8106/14819284.jpg

for the first part .. it is asking for ymax , let's think at ymax what happens to the final velocity? you are given yo=0 and t at max. height =3.5s .. can you go from there?


for the second part .. you are asked to find the height of the fence .. from the figure you can note that this height is the same height if you are aked to find y after 3s ..


for the third part .. you can notice from the figure that if you found the range of the travel (max. x the ball travels) then finding the difference between it and 95.8m , this way you find x .. but when it hits the ground what can you say about y??


I believe I gave you enough hints , please work or atleast try as much as you can and then show your detailed solution so if you got stuck we can help .. REMEMBER : members don't do your homework they just help :) ..
 
Last edited by a moderator:
Ok so I found Ymax by taking my Voy^2/2g. I then found the fences height by V^2=Voy^2+2gDeltaY. I am however, stuck on the last part, I know that Delta X is going to be the total distance minus the distance given in the problem of 95.8m. But in order to find total distance I need t total but when I calculate ttotal=Vfy-Voy/Ay I get an answer of 3.5s which would mean the ball didn't make it to the fence which is incorrect?
 
looking at the figure you can notice that there is some kind of symmetry, it should be obvious that when it reaches ymax. At t=3.5s it covers half of the distance of travel .. So from there can you find what should be the total time of the travel ??
 

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