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![[Post New]](/templates/default/images/icon_minipost_new.gif) 23 Nov 2007 23:38:31 IST
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ok guys here's another one a small ball thrown at an initial velocity v at an angle A to the horizontal strikes a vertical wall moving towards it at a speed v' and is bounced to the point from which it was thrown..Determine the time t from the beginning of the motion to the moment of impact neglecting friction losses. again the same conditions...though this time ill not give salutes unless i have the 100% correct ans...once again solution by tomorrow...
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:23:20 IST
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Time from initiation of motion to hit the wall = vSinA ( 2v ' + vCosA) / g(vCosA + v ' )
Will post the soln if its right. and its the time to hit the wall you are asking right ?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:27:34 IST
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that's correct...dont post the solution...and frm my side this challenge ends here...those who wud still like to try this problem may still try it....seems it was'nt as gud as the last one.so then another one?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:29:31 IST
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Let t = time to hit the wall t ' =time taken to come back to original point after hitting the wall t + t' = 2vSinA/g
V = Vel after hitting the wall As friction is not to be considered, e =1 (Final Vel of approach or separation ) / (Initial Vel of approach or separation )=1
So, V = 2v' + vCosA
Distance from initial point to wall = Distance from wall to initial point vCosA x t = V x t' So, vCosA x t = (2v' + vCosA)(2vSinA/g - t)
So, t = vSinA ( 2v ' + vCosA) / g(vCosA + v ' )
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:30:27 IST
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Heh , yeah man. Another one. Just keep it coming.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:31:52 IST
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ok just resolve this one issue and ill give u another 2 points...why did u take total time = time of flight?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:34:57 IST
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Because the wall exerts force/impulse only along x axis which is the direction perpendicular to the walls surface. and time of flight is decided by the vel of ball along y axis which remains unaltered since no force being exerted anytime by the wall along x axis.
So, along y axis. Initial vel is VSinA. So, time taken to come back to the ground will be same even if there was no wall.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Nov 2007 00:40:28 IST
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well that makes the solution complete....check out the nxt one
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