physics
posted by Billie .
A hot air ballon is ascending at 12m/s and at 80 m above the ground, a package is dropped over the sdie. How long does it take the package to hit the ground?
I can't figure out which equation to ues. but I know that v=12m/s, d=80m and t=?
Since the object is not acclerating I 'think' you would just use
s=(1/2)g*t^2 where s=80m
The object would have constant velocity until it's dropped. Then it's under the influence of gravity completely.
Check my reasoning on this.
I'm now quite sure I gave the wrong answer to this question.
The ballon's velocity is 12m/s upward and it's 80 m above the ground. A package is dropped over the side, and we want to know how long does it take for the package to hit the ground.
We should use
(1) s=(1/2)g*t^2 + v_o*t + x_o and
(2) v=g*t + v_o
Here x_o = 80m and v_o=12m/s. First we should determine how long it takes for the object to stop moving upward. Using (2) we find
0=9.8m/s^2*t + 12m/s so t=12/9.8s=1.22s. using this in (1) we get
s=(1/2)9.8m/s^2*(1.22s)^2+12m/s*1.22s + 80 = something you can do.
This is the height from which the stone in under the influence of gravity completely
Use that height and solve for t using
s=(1/2)g*t^2
Add that time to the 1.22s it needed going up to answer your question.
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