To solve the problem, it is necessary to use the concepts of gravitational force, centripetal force and trigonometric components that can be extrapolated from the statement.
By definition we know that the Force of Gravity is given by
![F_g=mg](https://tex.z-dn.net/?f=F_g%3Dmg)
Where,
m= Mass
g = Gravitational Acceleration
The centripetal force is given by,
![F_c = \frac{mv^2}{R}](https://tex.z-dn.net/?f=F_c%20%3D%20%5Cfrac%7Bmv%5E2%7D%7BR%7D)
Where,
m = Mass
v = Velocity
R = Radius
For the case described in the problem, the Force of gravity the net component would be given by sin?, While for the centripetal force the net component is in the horizontal direction, therefore it corresponds to the ![cos\theta](https://tex.z-dn.net/?f=cos%5Ctheta)
Then,
![F_g = mg sin\theta](https://tex.z-dn.net/?f=F_g%20%3D%20mg%20sin%5Ctheta)
![F_c = \frac{mv^2}{r}cos\theta](https://tex.z-dn.net/?f=F_c%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7Dcos%5Ctheta)
From the radius we have its length but not the net height, which would be given by
![r = L sin\theta](https://tex.z-dn.net/?f=r%20%3D%20L%20sin%5Ctheta)
So equating the equations we have to
![F_g = F_c](https://tex.z-dn.net/?f=F_g%20%3D%20F_c)
![mg sin\theta=\frac{mv^2}{r}cos\theta](https://tex.z-dn.net/?f=mg%20sin%5Ctheta%3D%5Cfrac%7Bmv%5E2%7D%7Br%7Dcos%5Ctheta)
![mg sin\theta=\frac{mv^2}{Lsin\theta}cos\theta](https://tex.z-dn.net/?f=mg%20sin%5Ctheta%3D%5Cfrac%7Bmv%5E2%7D%7BLsin%5Ctheta%7Dcos%5Ctheta)
Re-arrange to find v,
![v = \sqrt{\frac{gLsin^2\theta}{cos\theta}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BgLsin%5E2%5Ctheta%7D%7Bcos%5Ctheta%7D%7D)
Replacing with our values
![v = \sqrt{\frac{(9.8)(22*10^{-2})(sin^2 24)}{cos24}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B%289.8%29%2822%2A10%5E%7B-2%7D%29%28sin%5E2%2024%29%7D%7Bcos24%7D%7D)
![v = 0.624](https://tex.z-dn.net/?f=v%20%3D%200.624)
Therefore the tangential velocity of the mass is 0.624m/s