Answer:
Both spheres hit the ground at the same time, but sphere A lands twice as far as sphere B from the base of the tower.
Explanation:
Since both spheres are released from the same height, therefore, the displacement is the same to mean they hit the ground at the same time. Considering that the speed of ball A is double the speed of ball B, it implies that A will land twice as far as B from the base of the tower.
Both spheres hit the ground at the same time and at the same distance from the base of the tower.
The given parameters;
The time required for each sphere to reach the ground is calculated as;
[tex]h = v_0_y + \frac{1}{2} gt^2\\\\[/tex]
where;
Both spheres have zero initial vertical velocity
The time required for the two spheres to hit the ground is calculated as;
[tex]h = 0 + \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t_A = \sqrt{\frac{2h}{g} } \\\\t_B = \sqrt{\frac{2h}{g} } \\\\t_A = t_B[/tex]
The time required for both spheres to reach the ground is the same and the distance traveled is the same.
Thus, we can conclude that, both spheres hit the ground at the same time and at the same distance from the base of the tower.
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