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

Let the initial velocity at the lowest point be $v_0 = \sqrt{10gl}$.

Using the principle of conservation of mechanical energy between the lowest point and the horizontal position:

$\dfrac{1}{2}mv_0^2 = \dfrac{1}{2}mv^2 + mgl$

Substituting the value of $v_0$:

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

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

$v^2 = 8gl$

At the horizontal position, the tension in the string provides the necessary centripetal force. The weight of the bob acts vertically downwards and has no component along the horizontal string.

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

Substituting $v^2 = 8gl$:

$T = \dfrac{m(8gl)}{l} = 8mg$