Answer:
≈ 6.68 m/s
Explanation:
A suitable formula is ...
vf^2 -vi^2 = 2ad
where vi and vf are the initial and final velocities, a is the acceleration, and d is the distance covered.
We note that if the initial launch direction is upward, the velocity of the ball when it comes back to its initial position is the same speed, but in the downward direction. Hence the problem is no different than if the ball were initially launched downward.
Then ...
vf = √(2ad +vi^2) = √(2·9.8 m/s^2·1.0 m+(5 m/s)^2) = √44.6 m/s
vf ≈ 6.68 m/s
The ball hits the ground with a speed of about 6.68 meters per second.
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We assume the launch direction is either up or down.
