58)

conservation of energy

U1+K1=U2+K2

K1=K2=0

thsu U1=U2

hence h needs 2 b d same

b) conservation of energy

U1+K1=U2+K2

1/2mv^2=mgl

thus v=

l=2r

thus v2=

then again conservation of energy

1/2m(v2)^2=mg(r+rcos@)+1/2m(v3)^2

4mgr/2=mg(r+rcos@)+1/2*mgrcos@                      mgcos@=m(v3)^2/r    thus (v3)^2=grcos@

thus solvin u get cos@=2/3

height aboove d lowest point=r+2/3r