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A balloon starts rising from ground from rest with an upward acceleration 2 m/s^2. Just after 1 second a stone is dropped from it. Calculate the time taken by the stone to touch the ground?

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2 years ago

```							When the stone is dropped its initial velocity is the same as the velocity of the balloon at that instant The upward acceleration of the balloon is its resultant acceleration ... a = 2 m/s² balloon starts from rest ... so v(i) = 0 m/s balloon rises for 1 s b4 the stone is dropped ... so v(f) = v(i) + at v(f) = 0 + 2m/s² * 1s = 2 m/s vertically up so the initial velocity of the stone v(i)(stone) as it begins to fall = 2 m/s vertically up also v(f)² = v(i)² + 2as s = [v(f)² - v(i)²] / 2a s = [4 - 0] / 4 s = 1 m so assuming the stone is dropped from the bottom of the basket, the stone is dropped from a height of 1 m As soon as the stone is released ... its acceleration is just due to gravity b/c once free of the balloon the only force acting on it is gravity (ignoring air resistance) so for the stone ... take down as the positive direction: v(i) = -2 m/s a = 9.8 m/s² s = 1m s = v(i)t + (1/2)at² 1 = -2t + (1/2) * 9.8 t² 1 = -2t + 4.9t² 4.9t² - 2t - 1 = 0 t = [2 ± √(4 + 19.6)] / 9.8 t = [2 ± √23.6] / 9.8 so t = [2 + √23.6] / 9.8 ... [b/c the time taken is positive so ignore the negative possibilit
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2 years ago
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