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If two objects at rest and at the same height begin a free fall one second apart how long after the first object begins to fall will the two objects be 10 meters apart?
Wiki User
∙ 16y ago
The answeer will be between one and two seconds. The general formula for rectilinear motion is d = d0 + v0t + (1/2)at2,where d0 and v0 are initial displacement and initial velocity, respectively. For this problem, there is no initial displacement and zero initial velocity, so d0 = v0 = 0. Hence, the formula for calculating displacement in this problem is d = (1/2)at2.
To further simplify the math, we will assume that displacement from the initial position -- that is, how far the objects have fallen -- is a positive quantity, so we can think of the acceleration due to gravity as positive, as well. The formula for the displacement of the first object, therefore, is d1 = (1/2)at2. And the formula for the displacement of the second object, which is released one second later, is d2 = (1/2)a(t-1)2. Therefore, all we must do is solve for twhen d1 - d2 = 10. So, we have
(1/2)at2 - (1/2)a(t-1)2 = 10. t2 - (t-1)2 = 20/a t2 - (t2 - 2t +1) = 20/a 2t - 1 = 20/a t - 1/2 = 10/a t = 10/a + 1/2 If a = 9.8, then t = 1.52 s. This answer checks out because d1(1.52) = 11.32 meters, and d2(1.52) = 1.32 meters.
Wiki User
∙ 16y ago
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