A stone is dropped from a balloon rising upwards with velocity of 16 m/s. The stone reaches the ground in 4s. Calculate the height of the balloon when stone was dropped.

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Solution

$Suppose t_{1} is the time taken to reach to the maximum height and t_{2} is the time taken to come to ground from maximum height. Now we know that v = u - g t = 0 o r t = \frac{u}{g} = \frac{16}{9.8} = 1.63 s Now the body has travelled tor the ground in (4-1.63)=2.37 seconds So using the formula, total height h is h = \frac{1}{2} g t^{2} = \frac{1}{2} \times 9.8 \times 2 . 37^{2} = 27.52 m . Again th upward distance travelled by the balloon is h_{1} = u t - \frac{1}{2} g t^{2} = 16 \times 1.63 - \frac{1}{2} \times 9.8 \times 1 . 63^{2} = 13.06 m So the height of the balloon is (27.52 - 13.06) = 14.46 m$

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