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The only trajectory problem on my whole SI sheet (trajectory) I cannot do! 
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#1
Mar1012, 01:47 AM

P: 14

1. The problem statement, all variables and given/known data
This a trajectory problem. A ball is launched from ground level at a unknown angle. ( ∅_{o}= ? ) at an unknown velocity ( V_{o} = ? ) The ball travels 20 meters in the x direction and 25 meters up in the y direction. At this point (20,25) the ball hits a wall and a 45 angle below the horizon. Find the initial velocity ( Mag and direction ) and the speed when the ball hits the wall 2. Relevant equations the only equations we have been given are V_{yfinal} = V_{o}( Sin ∅_{o})  gt Δ Y = V_{o}( Sin ∅_{o})  1/2g (t)^2 ΔX = V_{o}(cos ∅) (t) V_{o}( Cos ∅_{o}) = Vf(Cos ∅_{o}) Xmax = (V_{o}^2 * (Sin 2∅))/ g 3. The attempt at a solution I cannot post all my notes. I have finished, and double checked all my other answers on 2 whole trajectory work sheets. But I cannot for the life of me figure out how to find the initial angle as a function of the final angle again the change in distance. I have even tried switching the whole problem around and shooting at 45° from 25 meters up and I still cannot find a result that works. I have done probably 1011 different tries of canceling out different variables (especially Vo and Vf and T) and finding them as products or sums of other variables, but I can never find a way to get the original y velocity or the orignal x velocity. I do know that when the ball hits the wall the Y velocity is = to the (x velocity), because tan1 45° = 1 (y/x) I have spent over 2 hours on this problem and I have searched the interwbs for help but I cannot for the life of me figure out how to find the original angle if I know the final angle but don't know either the final or intial x or y velocities. I also don't know the time I really, really don't like asking for answers to problems because I really like sitting down with a cup of joe and figuring it out for myself and actually learning, but I literally have grown some grey hairs and ruined my weekend over the amount of stress I have experienced over this problem, and I would really really appreciate if someone could answer this one if not give me a HUGE hint 


#2
Mar1012, 04:47 AM

P: 25

If the ball hits the wall at an angle of 45° after travelling 25m upwards, what does that tell you about what the maximum height reached would have been if the wall was not present?



#3
Mar1012, 04:54 AM

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P: 5,363




#4
Mar1012, 05:46 AM

HW Helper
P: 2,318

The only trajectory problem on my whole SI sheet (trajectory) I cannot do!
The second derivative anywhere is 9.8 Surely if you put all that together the answer can be extracted That enables you 


#5
Mar1012, 02:18 PM

P: 14




#6
Mar1012, 02:20 PM

P: 14

"The parabolic path followed passes through 0,0 and 20,25, and the gradient at (20,25) is 1
The second derivative anywhere is 9.8 Surely if you put all that together the answer can be extracted That enables you" I just started calculus 2 weeks ago. I am decent with derivatives but to be honest I have never done a lick of integration, which is what I would need to go from having the derivative to the original right? 


#7
Mar1012, 03:27 PM

P: 628

Define a time, t, for yourself, that being how long it takes to get to the wall. So we know at time t its horizontal and vertical velocities are both 20/t.
Now ask a different question  with initial velocities 20/t both horiz and vertical, from a height 25, what angle will its flight intersect the horizontal if it goes from there? 


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