Aball thrown vertically upward with an initial speed of 5 m/s, what will its speed be when it returns to its starting point.(in the absence of air resistance)

The answer is indeed 5m/s.  The easiest way to understand this is conservation of energy.  The kinetic energy put into the ball at the beginning of its trajectory is KE=(1/2)mv², where m is the unknown mass of the ball.  At the very top of its flight it stops and turns around and starts to fall back down.  At the moment it stops all of its energy is gravitational potential energy, given by
PE=mgh.  The conservation of energy says this potential energy is exactly equal to the initial kinetic energy.  The energy is changing forms as the ball rises and falls.  At the beginning when the ball is released and its height is still at the starting position, we can say the PE is 0 and the KE is maximum.  As the ball goes up its height, h, increases, increasing the potential energy PE=mgh. As this happens though, the ball is slowing down and its kinetic energy is decreasing.  At the peak it is all PE.  When it finally returns to the starting position, all of its original energy is now kinetic, meaning it's speed is what it was at the beginning, 5m/s.  At any point in the trajectory, the sum of the PE and KE must add up to the total you started with.

The speed was not exceeded so it was slow