Explanation
When the string has traversed an angle $$\theta$$ , then
$$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ l }$$ , and conservation of energy gives
$$\dfrac { m{ u }^{ 2 } }{ 2 } =\dfrac { m{ v }^{ 2 } }{ 2 } +mgl(1-cos\theta ).$$ then the string becomes slacked tension in the string becomes zero. Solving the above two equations to get the value of $$\theta$$
$$\Rightarrow cos\theta =\dfrac { -1 }{ 3 } and { v }^{ 2 }={ u }^{ 2 }-2gl(1-cos\theta )=\dfrac { gl }{ 3 } $$ from that point the maximum height reached is $$\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g }$$ maximum height $$= l(1-cos\theta )+\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g } =\dfrac { 4l }{ 3 } +\dfrac { 4l }{ 27 } =\dfrac { 40l }{ 27 } $$
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