Question

0
A sandbag ballast is dropped from a balloon that
ascending with a velocity of 40 ft s. If the sandbag reaches
the ground in 20s, how high was the balloon when the ba
was dropped? Neglect air resistance.​

Answer 1

Answer:

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Answer 2

6400 ft

Explanation:

Initial velocity of balloon=u=40 ft/s

Time=20 s

Initial velocity of bag=u=-40 ft/s

Because sandbag and balloon are moving in opposite direction.

We know that

$s=ut+\frac{1}{2}gt^2$

Where $g=32 ft/s^2$

Using the formula

$s=(-40)(20)+\frac{1}{2}(32)(20)^2$

$s=5600 ft$

Distance traveled by balloon=$s'=velocity\times time=40\times 20=800ft$

Total distance =5600+800=6400 ft

Hence, the height of balloon when the bag was dropped=6400 ft

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