Question

A hot air balloon is traveling vertically up-
ward at a constant speed of 2.3 m/s. When
it is 19 m above the ground, a package is
released from the balloon.
After it is released, for how long is the
package in the air? The acceleration of gravity
is 9.8 m/s2 .
Answer in units of s

Answer 1

After it is released, for how long is the package in the air?

Once released, the package is in free fall with an acceleration of
a = - g = - 9.8 m/s2
We know its initial velocity and initial position

vi = 2.3 m/s
yi = 15 m

The later position of the package is given by

y = yi + vi t + (1/2) a t2
y = 15 m + (2.3 m/s) t + (1/2) ( - 9.8 m/s2) t2

Now we set y = 0 and solve for t. Right?
0=15+2.3t-4.9t^2 which is -1.53 & 2 but the answer isnt correct. What am I doing wrong? Please help?

Answer 2

Am I using the right formula?

Answer 3

still reading - give me a sec...

Answer 4

yi should be 19m not 15m

Answer 5

therefore 2.22

Answer 6

freak yea. ugh.

Answer 7

i lost two points for that stupid error....

Answer 8

Thank YOU!!!!

Answer 9

:) at least you learnt something - just like I did ...

Answer 10

yw