Question

If a hot air balloon is travelling vertically upward at a constant speed of 2.3 m/s. When it is 19 m above the ground, a package is release from the balloon.

Answer 1

After it is released, for how long is the package in the air? The acceleration of gravity is 9.8 m/s^2 . Answer in units of seconds.
What is its speed just before the impact with the ground? Answer in units of m/ second.

Answer 2

Now assume the hot air is moving vertically downward at a constant speed of 2.3 m/s. After the package is released, how long is it in the air? Answer in seconds.
What is its speed just before impact with the ground? Answer in m/s.

Answer 3

apply formula v=u + at

Answer 4

ok

Answer 5

jst plug in all parameter u ll get "t"

Answer 6

is u the distance?

Answer 7

yea...

Answer 8

Does 1.7 seconds sound right?

Answer 9

it doesnt to me. :(

Answer 10

i didn't check ans .....bt what is ans.....exactly...then..

Answer 11

srry , apply this formula
s=ut+1/2 at^2 ok

Answer 12

here s= distance

Answer 13

I'm not for sure... I set it up: 2.3 m/s = 19m +9.8m/s^2 (t)
2.3 m/s - 19m =9.8m/s^2 (t)
(2.3 m/s)/(9.8 m/s^2) - (19m)/(9.8m/s^2) = t

Answer 14

ok 1 second.

Answer 15

am i using quadratic on this?

Answer 16

jst pulg in all values r given ...

Answer 17

its quadratic in t

Answer 18

We know the distance the package falls, but just after releasing the package is initially traveling upwards. We can use the following equation to solve for \(t\):\[d = v_ot - {1 \over 2} gt^2\]where \(d = d_o + {v_o^2 \over 2g}\), \(v_o\) is the velocity of the balloon, \(d_o\) is the height of the balloon when the package is dropped, and \(g\) is gravitational acceleration.
We can use the following equation to solve for the velocity at impact with the ground:\[v =v_o - g t\]where \(v_o\) is the velocity of the balloon.

Answer 19

Ok so the distance = 18.27(using the formula) if i put that into a quadratic equation it is not a "real solution"

Answer 20

The highest the box achieves while traveling upwards should be higher than 19m because the box is initially travelling upwards. Therefore, gravity must first slow the box to zero velocity, then begin accelerate it towards the surface of the earth. The box simply accelerates downward when the balloon travels downward. In this case the distance is simply 19m.