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Answer 1 · 2018-02-03T05:34:02

Immediately after the launch,the body has a horizontal velocity of #8 m/s#,So its kinetic energy is #1/2 0.36* 8^2 J = 11.52 J#

And,potential energy at that point is #mgh# i.e #0.36*10*45 J=162 J#

So,total energy just after the launch is #(11.52+162) J =173.52J#.....a

Now,on moving from A to B it has come down by #(45-32) m=13 m#

So,decrease in potential energy will be #mg*13=46.8 J#...b

So,this much energy has been converted to kinetic energy,as total energy is fixed.

So,at point B, total kinetic energy is #(46.8+11.52)J = 58.32 J# (because horizontal component of velocity is unchanged).....c

Now, if it has a total kinetic energy of #58.32 J# at B so we can equate it with #1/2mv'^2# (#v'# is the speed at point B)

So,we get, #v'=18 m/s#.....d

When it will hit the ground,its total energy will be kinetic energy and will be equal to its initial energy i.e #173.52 J#....e