To solve this problem, we can use the equations of motion. In this case, since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. We'll assume the acceleration due to gravity is approximately 9.8 m/s².
Given: Initial vertical velocity (u) = 0 m/s Time taken (t) = 3.0 s Acceleration due to gravity (a) = 9.8 m/s²
Using the equation:
s = ut + (1/2)at²
where: s = displacement u = initial velocity t = time taken a = acceleration
Since the ball hits the ground, the displacement in the vertical direction (s) will be the distance traveled vertically downward. We can solve for s using the equation:
s = (1/2)at²
Substituting the known values:
s = (1/2)(9.8 m/s²)(3.0 s)² s = 44.1 m
Now, to find the final velocity (v), we can use the equation:
v = u + at
Since the ball was thrown horizontally, its horizontal velocity (u) remains constant throughout its motion. Therefore, the final horizontal velocity (v) will also be equal to the initial horizontal velocity, which is 10.0 m/s.
So, the speed of the ball just before it hits the ground will be 10.0 m/s (assuming the magnitude of the horizontal velocity remains constant).