# 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.