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

Let $u$ be the velocity at the lowest point and $v$ be the velocity at the highest point.

Using conservation of mechanical energy between the lowest and highest points:

$\dfrac{1}{2}mu^2 = \dfrac{1}{2}mv^2 + mg(2l)$

Given $u = \sqrt{10gl}$, substituting this yields:

$\dfrac{1}{2}m(10gl) = \dfrac{1}{2}mv^2 + 2mgl$

$5mgl = \dfrac{1}{2}mv^2 + 2mgl$

$\dfrac{1}{2}mv^2 = 3mgl \Rightarrow v^2 = 6gl$

At the highest point, both tension $T$ and weight $mg$ act downwards towards the center of the circular path. The equation for centripetal force is:

$T + mg = \dfrac{mv^2}{l}$

Substituting $v^2 = 6gl$:

$T + mg = \dfrac{m(6gl)}{l}$

$T + mg = 6mg$

$T = 5mg$