With what speed the ball was thrown up from the surface of the earth, if at a height of 1 meter above

With what speed the ball was thrown up from the surface of the earth, if at a height of 1 meter above the point of throwing its potential energy is 4 times greater than the kinetic energy. The potential energy on the earth’s surface is zero.

The potential energy of the ball at a height of h = 1 m is equal to mgh, where m is the mass of the ball, g is the acceleration of gravity.
The kinetic energy is four times more, i.e. equal to 4mgh.
The total energy is 4mgh + mgh = 5mgh.
Since the total energy does not change during the flight, then at the throwing point it is equal to 5mgh, and since the potential energy is equal to 0, then the total energy is equal to the kinetic one, i.e. mv ^ 2/2.
mv ^ 2/2 = 5 mgh
v ^ 2 = 10mgh / m = 10gh
v = √ (10gh) = √ (10 * 10 m / s ^ 2 * 1 m) = 10 m / s