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# Hotairballon is dropping at 2.2m/s when person drops camera. If camera drops at 40m above ground how long does it take camera to reach floor?

Also, what is its velocity just before it lands? (m/s) Let upward be positive direction

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- WhomeLv 71 month agoFavorite Answer
s = s₀ + v₀t + ½at²

0 = 40 +(- 2.2)t + ½(-9.8)t²

0 = 40 - 2.2t - 4.9t²

t = (2.2 ±√(2.2² - 4(-4.9)(40))) / (2(-4.9))

t = -3.09 s Ignore

or

t = 2.6414...

t = 2.6 s ANSWER

v = v₀ + at

v = -2.2 - 9.8(2.6)

v = - 28 m/s

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- 3 weeks ago
The camera reaches ground as a horizontal projectile, as such it’s time of flight should be

t = square root of ( 2h/g)= sq.root( 2x40/9.8)

= 20/7 sec = 3 sec (nearly)

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