To find the speed of the ball just before it hits the ground, we can use the fact that horizontal motion and vertical motion are independent of each other.
In horizontal motion, the ball maintains a constant velocity since there is no horizontal force acting on it. Therefore, the horizontal component of the initial velocity (10.0 m/s) remains constant throughout its flight.
In vertical motion, we can use the equation of motion for an object in free fall:
h = ut + (1/2)gt^2
where: h = vertical displacement (height) of the ball = 0 (since it hits the ground) u = initial vertical velocity of the ball = 0 (since it is thrown horizontally) g = acceleration due to gravity = 9.8 m/s^2 (assuming Earth's gravity)
We can rearrange the equation to solve for time (t):
t = sqrt(2h/g)
Substituting the values: t = sqrt(2 * 0 / 9.8) = 0 s
Since time (t) is 0, it means the ball takes no time to travel vertically and hits the ground instantaneously.
Therefore, the speed of the ball just before it hits the ground is the same as its horizontal velocity, which is 10.0 m/s.