Calculating Time of Impact for Released Ballast Bag in 2D Kinematics

[sportzmaniac]

Homework Statement

A hot-air balloon is rising straight up with a speed of 2.4 m/s. A ballast bag is released from rest relative to the balloon when it is 9.6 m above the ground. How much time elapses before the ballast bag hits the ground

known data

y direction
vo = 2.4 m/s
a = -9.8m/s
x = -9.6m

Homework Equations

v = vo + at
v-squared = vo-squared + 2ax
x = volt + 1/2a(t-squared)

The Attempt at a Solution

By plugging numbers into v-squared = vo-squared + 2ax, I got v = 13.9 m/s. Then, I plugged that and other data into v = vo + at, adn it came out to 1.17s (or 1.2s). However, I am entering this into a website which tells you whether you are right or wrong. It marked 1.2 wrong, and I only have one more guess. Is 1.17 right, or did I mess this up somehow?

 

[inhere]

I'm not sure if you're wrong, sometimes the programming behind those sites can be a bit shady, but it doesn't seem like you defined the final velocity as v=-13.9 m/s, where the initial velocity v = + 2.4 m/s. This will change your final answer.

 

[sportzmaniac]

I'm not sure if you're wrong, sometimes the programming behind those sites can be a bit shady, but it doesn't seem like you defined the final velocity as v=-13.9 m/s, where the initial velocity v = + 2.4 m/s. This will change your final answer.

Wow, I messed that up. However, when plugging that into v = vo+ at It came out to 1.66, which I plugged into the site as 1.6s, and it told me that was wrong. Anyone know what I did wrong?