To solve this problem, we can use the principles of projectile motion. We'll consider the motion of ball A and ball B separately.
For ball A:
- It is dropped from rest, so its initial velocity is 0 m/s.
- The final height above the ground where it needs to meet ball B is h/2.
For ball B:
- It is thrown vertically upwards from the ground.
- The final height above the ground where it needs to meet ball A is also h/2.
Let's determine the time it takes for each ball to reach the meeting point at a height of h/2 above the ground.
For ball A: We can use the equation for vertical displacement during free fall:
Δy = V₀t + (1/2)gt²,
where: Δy = vertical displacement of h/2, V₀ = initial velocity of 0 m/s, t = time taken to reach the meeting point, g = acceleration due to gravity (-9.8 m/s²).
Substituting the values:
h/2 = 0 m/s * t + (1/2)(-9.8 m/s²)t².
Simplifying the equation:
h/2 = (-4.9 m/s²)t².
Multiplying by 2 to remove the fraction:
h = -9.8 m/s² * t².
Simplifying further:
t² = -h / (9.8 m/s²).
Taking the square root:
t = sqrt(-h / (9.8 m/s²)).
Note that we consider the positive square root since time cannot be negative.
Now, for ball B: Since ball B is thrown vertically upwards, it experiences a negative acceleration due to gravity (-9.8 m/s²). At the highest point, its vertical velocity becomes 0.
We can use the equation for vertical velocity:
V = V₀ + gt,
where: V = final vertical velocity (unknown), V₀ = initial vertical velocity (unknown), g = acceleration due to gravity (-9.8 m/s²).
At the meeting point, ball B has reached its maximum height, so V = 0.
We can rewrite the equation as:
0 = V₀ + (-9.8 m/s²)t.
Solving for V₀:
V₀ = -(-9.8 m/s²)t.
V₀ = 9.8 m/s² * t.
Now, we know that the initial velocity of ball B at the meeting point (h/2) is 9.8 m/s * t.
Therefore, the speed of ball B at the meeting point (h/2) is 9.8 m/s * t.