To solve this problem, we can use the equations of motion for projectile motion. Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s. We can assume that the acceleration due to gravity is 9.8 m/s² downward.
We can use the equation: h = ut + (1/2)gt²
Where: h = vertical displacement (20 m) u = initial vertical velocity (0 m/s) g = acceleration due to gravity (-9.8 m/s²) t = time of flight
Since the ball is thrown horizontally, the time taken to reach the ground is the same as the time of flight. Rearranging the equation, we get:
t = sqrt(2h/g)
Substituting the values, we have: t = sqrt(2 * 20 / 9.8) ≈ 2.02 s
Now, we can use the formula for horizontal distance traveled: d = horizontal velocity × time of flight
Since the ball is thrown horizontally, the horizontal velocity is 5 m/s. Substituting the values, we have: d = 5 m/s × 2.02 s ≈ 10.1 m
Therefore, the ball will land approximately 10.1 meters from the base of the cliff.