Solving for Time: Dropped Pop Can Reaches the Ground

[powerxranger1]

Homework Statement

A hot air balloon is rising upward with a constant velocity of 4m/s. As the balloon reaches a height of 4m above the ground the balloonist accidently drops a can of pop over the edge of the basket. How long does it take the pop can to reach the ground?

V2 = 4m/s [up]
v1 = 0
d = 4m

Homework Equations

d=1/2 (v2 + v1) t

The Attempt at a Solution

I got the anser 2 seconds! Please tell me if i did it correctly

 

[alphysicist]

Hi powerxranger1,

Homework Statement

A hot air balloon is rising upward with a constant velocity of 4m/s. As the balloon reaches a height of 4m above the ground the balloonist accidently drops a can of pop over the edge of the basket. How long does it take the pop can to reach the ground?

V2 = 4m/s [up]
v1 = 0
d = 4m

Homework Equations

d=1/2 (v2 + v1) t

The Attempt at a Solution

I got the anser 2 seconds! Please tell me if i did it correctly

No, I don't believe that's right. The initial velocity of the can is 4m/s upwards, like you have, but the final velocity is not given. (The final velocity in this case would be the velocity just before the can touches the ground, so it would not be zero.)

So you know the initial velocity and the displacement (but check the signs!); what else do you know about the can's motion? That will help you choose the best kinematic equation to find the time.

 

[powerxranger1]

Hm, I'm not really sure that there is any other information given about the can, since there's only two numbers given. I don't understand how to find another piece of information.

 

[alphysicist]

Hm, I'm not really sure that there is any other information given about the can, since there's only two numbers given. I don't understand how to find another piece of information.

The can is in free fall; what does that tell you? That will give you the third quantity, and so you can find the time.

 

[powerxranger1]

Oh. Does that mean it's just 9.8 m/s?

 

[alphysicist]

Oh. Does that mean it's just 9.8 m/s?

Yes, that's the right magnitude for the acceleration. (And the units are m/s^2.) Just remember to use the correct signs for the initial velocity, displacement and acceleration when you use them in your equation.