A bag of ballast is accidentally dropped from a balloon which is stationary at an altitude of 4900m. How long does it take for the bag to hit the ground?

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A bag of ballast is accidentally dropped from a balloon which is stationary at an altitude of 4900m. How long does it take for the bag to hit the ground?

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Answer 1 · 2018-03-05T04:25:04

$31.6$ $31.6$ s

Explanation:

A good equation to know for calculating time of falls due to gravity alone is:

$h = \frac{1}{2} g t^{2} \to t = \sqrt{\frac{2 h}{g}}$ $h = 1/2 g t^2 -> t = sqrt((2h)/g)$

Here, we can use $h = 4900$ $h = 4900$ m. There is no gravity mentioned, so I will simply use $9.8$ $9.8$ m/ $s^{2}$ $s^2$ .

$t = \sqrt{\frac{(2) (4900)}{9.8}} = 31.6$ $t = sqrt(((2)(4900))/9.8) =31.6$ s

Answer 2 · 2018-03-05T04:45:50

Use the equations of motion to answer the question.

Explanation:

It is given that balloon is stationary at height $(s) = 4900 m$ $(s) = "4900 m"$ . The initial velocity $(u) = 0 m/s$ $(u) = "0 m/s"$ .

As it falls in the downward direction, it accelerates with an acceleration of $g = {9.81m/s}^{2}$ $g = "9.81m/s"^2$ (for simple calculations, $g = {9.8 m/s}^{2}$ $g = "9.8 m/s"^2$ )

So

$4900 = 0 \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $4900 = 0*t + 1/2*g*t^2$

On solving

$t = \sqrt{\frac{2 \cdot 4900}{9.8}}$ $t = sqrt((2 * 4900)/9.8)$

$t = \sqrt{1000} s$ $t = sqrt(1000) " s"$

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A bag of ballast is accidentally dropped from a balloon which is stationary at an altitude of 4900m. How long does it take for the bag to hit the ground?

2 Answers

Explanation:

Explanation:

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