How Far Does a Sand Bag Fall from a Rising Hot Air Balloon After Five Seconds?

[petern]

A hot air balloon is rising at 5.0 m/s when a bag is released. Five seconds later, how far below the hot air balloon is the sand bag. Answer: 122.5 m.

So I use the equation (1/2)(a)(t)^2 to figure out the distance the bag has traveled. I use the equation -Vt to figure out how far the balloon has risen. I end up getting 97.5 m when I combine the 2 but when I use just (1/2)(a)(t)^2, I get the correct answer.. Can someone please explain to me what I'm doing wrong?

 

[Midy1420]

did you take into account that when the sandbag begins to fall it has an upward velocity of 5m/s?

 

[petern]

did you take into account that when the sandbag begins to fall it has an upward velocity of 5m/s?

Wow! I can't believe I forgot about that. Thanks a lot.

 

[Chi Meson]

The actual equation is $$d=v_o t + 1/2at^2$$

 

[petern]

The actual equation is $$d=v_o t + 1/2at^2$$

Thanks!