Question

Can you help with this?

Answer 1

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