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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
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tree449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH12/EX12.4/12_4.sce
parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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+//Crossing of balls
+//refer fig. 12.8
+//1.for motion of first ball
+au=0
+//1s=30-h
+aa=9.81 //m/sec^2
+//2.for motion of second ball
+bu=15 //m/sec
+//s=h
+ba=-9.81 //m/sec^2
+//30-h=0*t+(9.81*t^2)/2 ...(1)
+//h=15*t-(9.81*t^2)/2 ...(2)
+//solving (1) and (2)
+t=30/15
+h=15*2-(9.81*2^2)/2 //m
+//at t=2
+//downward velocity of first ball
+v1=0+9.81*2 //m/sec
+//Upward velocity of second ball
+v2=15-9.81*2 //m/sec
+//relative velocity vr
+vr=v1-(-v2) //m/sec
+printf("\nt=%.2f sec\nh=%.2f m\nvr=%.2f m/sec",t,h,vr)
+