θ ખૂણાવાળા ઢાળનો ઉપરનો અડધો ભાગ ઘર્ષણરહિત

For upper half

${v^2} = {u^2} + 2al/2 = 2(g\sin \theta )l/2 = gl\sin \theta $

For lower half

$⇒$ $0 = {u^2} + 2g(\sin \theta - \mu \cos \theta )\frac{l}{2}$

$⇒$ $ - gl\sin \theta = gl(\sin \theta - \mu \cos \theta )$

$⇒$ $\mu \cos \theta = 2\sin \theta \Rightarrow \mu = 2\tan \theta $