The height below the tower at which coin 1 pass coin 2 is 89.04 m.
The given parameters:
height of the tower, h = 100 m
initial velocity of coin 1, v = 15 m/s
time spent in air by coin 1 before coin 2 was dropped = 2s
To find:
Find the maximum height attained by coin 1 before falling to the ground:
[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]
Find the time taken for coin 1 to fall to the ground:
Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m
[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]
But the time taken for the coin 1 to reach 11.25 m above the tower:
[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]
Total time spent by coin 1 before reaching ground with respect to coin 2:
time = (1.5 s + 4.72 s) - 2 s
time = 4.22 s
Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.
Find the total time taken for coin 2 to fall to the ground:
Height of coin 2 above the ground = 100 m
Total time taken by coin 2 before falling to the ground is calculated as:
[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]
The time at which coin 1 will pass coin 2 is 4.22 s.
Find the height below the tower when the time is 4.22 s.
[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]
Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.
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