The bob of mass m of a resting pendulum is given a sharp hit,...

Let the lowest point of the pendulum be the reference level for potential energy.

Initial kinetic energy is given by:

$K_i = \dfrac{1}{2} m u^2 = \dfrac{1}{2} m (10gl) = 5mgl$

When the string makes an angle of 60^{}^{{}^{\circ}} with the upward vertical, it makes an angle of \theta = 180^{}^{{}^{\circ}} - 60^{}^{{}^{\circ}} = 120^{}^{{}^{\circ}} with the downward vertical.

The height of the bob from the lowest point is:

h = l - l \cos\theta = l(1 - \cos 120^{}^{{}^{\circ}}) = l(1 - (-0.5)) = 1.5l

Using conservation of mechanical energy:

$K_i + U_i = K_f + U_f$

$5mgl + 0 = \dfrac{1}{2} m v^2 + mg(1.5l)$

$\dfrac{1}{2} m v^2 = 3.5mgl \Rightarrow v^2 = 7gl$

The equation of motion for the bob in the radial direction is:

$T - mg \cos\theta = \dfrac{mv^2}{l}$

Substituting the values of $\theta$ and $v^2$:

T - mg \cos 120^{}^{{}^{\circ}} = \dfrac{m(7gl)}{l}

$T - mg(-0.5) = 7mg$

$T + 0.5mg = 7mg$

$T = 6.5mg$