Calculating Motion of a Test Balloon on a Methane Planet

[robax25]

should I use terminal velocity equation?

 

[robax25]

I do not consider Terminal velocity. So total net force Fnet = mg-Fb= 4kg*0.425 m/s² - (0.17N+42.5N)= -40.97N so ma= -40.97N; a= -40.97N/4kg= -10.24 m/s²

 

[haruspex]

-40.97N/4kg

How much mass is accelerating?

 

[robax25]

Fnet = mg-Fb= 19kg*0.425 m/s² - (0.17N+42.5N)= -34.595N a=-34.595N/19kg=-1.8 m/s²

 

[haruspex]

Fnet = mg-Fb= 19kg*0.425 m/s² - (0.17N+42.5N)= -34.595N a=-34.595N/19kg=-1.8 m/s²

Looks better.
Why the minus sign? Are you taking down as positive?

 

[robax25]

yes but it is not a matter

 

[robax25]

Now i need to find out the time to fall

 

[haruspex]

Now i need to find out the time to fall

Fall? Do you mean time for the box to reach the surface?
Since you are ignoring drag, you can use the relevant equation you posted in post #1.

 

[robax25]

yes

 

[robax25]

y=0.5 at² = 0.5* 1.8m/s²*t²
t²= 10m/(0.5*1.8m/s²)=11.11s²
t=3.33s.

 

[haruspex]

y=0.5 at² = 0.5* 1.8m/s²*t²
t²= 10m/(0.5*1.8m/s²)=11.11s²
t=3.33s.

Looks right

 

[robax25]

the graph would be like that but I need to change the value.

 

[haruspex]

Yes.

the graph would be like that but I need to change the value.