Grade 12Mechanics
4th one, i have tried many times but not able to solve
1 Answer
Bhavesh Mehta
Approved Tutor Answer9 Years agoFirst, by using conservation of energy, find out the velocity at O. U^2= 2gH1=>U^2= 300.Then use conservation of energy again to find out velocity at B. 1/2mV^2=1/2mU^2 - mgH2.Putting the values of U^2 from above(300) and H2=10m we get V^2 = 100. So V=10.From then on the ball follows a parabolic trajectory of which maximum height can be found out as H= V^2 sin^2x/ 2g. The angle at which it leaves is the same as slope of track.Sin37°= 3/5. The height comes out to be 3.6m.Adding this to the height of 10m that it achieved earlier before leaving the track totals 13.6m.
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