Question

A heavy particle hanging from a fixed point by alight inexetendable string of length l is projected horizontly with speed gl

Answer 1

initially total energy (TE) = mu2/2 + PE = mu2/2 (PE of ground is taken as 0)

=mu2/2

when it sustends @ with verticle then let its velocity is v ...

finally total energy = mv2/2 + PE

PE = mgR(1-cos@) , so total energy finally

TE = mv2/2 + mgR(1-cos@) .............2

since total energy is conserved so eq1 = eq2

mv2/2 + mgR(1-cos@) = mu2/2 ................3

now let tention in string at this pos is T then

T - mgcos@ = mv2/R

T = mg (given) , so

mv2 = (mg-mgcos@)R ............4

solving eq 4 & 3

3(1-cos@) = 1

cos@ = 2/3

@ = cos-1(2/3)