Since the ball is thrown horizontally, there is no initial vertical velocity component. The only force acting on the ball in the vertical direction is the force of gravity. As the ball falls, it accelerates downward at a rate of approximately 9.8 m/s².
To find the time it takes for the ball to hit the ground, we can use the equation:
y = y0 + v0y × t + (1/2) × a × t²
where: y is the vertical displacement (final position) = -20 m (negative because it's below the starting point) y0 is the initial vertical position = 0 m v0y is the initial vertical velocity component = 0 m/s a is the acceleration due to gravity = -9.8 m/s² (negative because it's downward) t is the time taken to reach the ground (what we want to find)
Plugging in the values, we have:
-20 m = 0 m + 0 m/s × t + (1/2) × (-9.8 m/s²) × t²
Simplifying the equation:
-20 m = -4.9 m/s² × t²
Dividing both sides by -4.9 m/s²:
t² = 20 m / (4.9 m/s²) t² ≈ 4.08 s²
Taking the square root of both sides:
t ≈ √(4.08 s²) t ≈ 2.02 s
So, it takes approximately 2.02 seconds for the ball to hit the ground. Now, we can find the magnitude of the vertical velocity component when the ball hits the ground by using the equation:
v = v0 + a × t
where: v is the final velocity (vertical component) v0 is the initial velocity (vertical component) = 0 m/s a is the acceleration due to gravity = -9.8 m/s² (negative because it's downward) t is the time taken to reach the ground = 2.02 s
Plugging in the values, we have:
v = 0 m/s + (-9.8 m/s²) × 2.02 s v ≈ -19.8 m/s
Therefore, the magnitude of the vertical velocity component when the ball hits the ground is approximately 19.8 m/s.