Projectile Motion of a Baseball Problem

AI Thread Summary
A baseball is hit and reaches its maximum height 3.5 seconds after being struck, then takes an additional 3.0 seconds to clear a fence located 95.8 meters away. The discussion involves using kinematic equations to calculate the maximum height of the baseball, the height of the fence, and the distance it travels beyond the fence. Participants emphasize the importance of understanding the symmetry of projectile motion and the relationship between time and distance traveled. There is a focus on solving for total time of flight to determine how far beyond the fence the ball lands. The conversation encourages users to attempt calculations and share their work for further assistance.
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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 :) ..
 
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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 ??
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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