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Step-by-Step Solution
Step 1: Identify the Motion
Both balls A and B are projected vertically from the same height: A upward and B downward. They have the same initial speed (magnitude) but in opposite directions. Let the magnitude of the initial velocity be $u$ and the height of the building be $h$.
Step 2: Set Up Kinematic Equations
Under constant acceleration due to gravity $g$, the relation between initial velocity $u$, final velocity $v$, displacement $s$, and acceleration $a$ is given by the kinematic equation:
$v^2 = u^2 + 2as.$
Here, $a = g$ (taking downward as the positive direction), and the displacement $s = h$ is the same for both balls from the point of projection to the ground.
Step 3: Apply the Equation to Each Ball
For Ball B (thrown downwards):
Its initial velocity is $+u$ (since it is thrown downward and downward is considered positive).
Hence, $v_B^2 = u^2 + 2gh.$
For Ball A (thrown upwards):
Its initial velocity is $-u$ (because it is thrown upward but we are considering downward as positive).
Hence, $v_A^2 = (-u)^2 + 2gh = u^2 + 2gh.
In both cases, the final velocity squared turns out to be the same: $u^2 + 2gh$.
Step 4: Compare Final Velocities
Since $v_A^2 = v_B^2$, their magnitudes are equal. This means:
$v_A = -v_B$ in direction, but $|v_A| = |v_B|.$
Thus, the speeds (magnitudes of velocity) of the two balls on reaching the ground are equal. This does not depend on their masses.
Final Answer
The correct choice is ${v_A} = {v_B}$.
