(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A hot air balloon is traveling vertically upward at a constant speed of 4.4 m/s. When it is 27 m above the ground, a package is released from the balloon.

The acceleration of gravity is 9.8 m/s^2.

a) After it is released, for how long is the package in the air.

b) What is its speed just before impact with the ground?

2. Relevant equations

x = x0 + v0t + 1/2at^2

v = v0 + at

3. The attempt at a solution

a)

0 = 27 + 4.4t + 1/2(-9.8)t^2

t = 2.839 s ***this answer is wrong.

Is it wrong because v0 should equal 0 and not 4.4?

b)

v = 4.4 - 9.8(2.839)

v = -23.42 m/s ***this answer is obviously wrong because t is wrong.

But, should v0 be zero in this case as well?

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# Homework Help: Simple Kinematics

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