Question

A hot-air balloon is ascending at the rate of 13 m/s and is 90 m above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

Answer 1

\[v ^{2}-u ^{2}=2as\].... use the formula to find final velocity nd with the help of that and using the acceleration formula find out the time taken..

Answer 2

check if the answer is correct...nt too sure...but should be correct

Answer 3

dont have the answer ... so final velocity is the speed with which it hits the ground?

Answer 4

yes...final velocity will be the speed with which it hits the ground...

Answer 5

can u tell me what answers do u get?

Answer 6

i got final velocity around 40m/s
nd time should be around 2.7 secs

Answer 7

yes thats what i got but this is wrong :/

Answer 8

what should the answer be??

Answer 9

i dont know, it only tells me if the answer is correct or incorrent :(

Answer 10

does it show time as 15.4 secs??

Answer 11

it does NOT give me answers :/

Answer 12

??

Answer 13

but it does tell u whether it is correct or no...check whether 15.4 is correct

Answer 14

how did u get 15.4s ?

Answer 15

by appling the formula \[s=ut+1/2at ^{2}\]....its just a try

Answer 16

First you need to figure out how high the package is when it starts falling back to Earth:
\[d=\frac{V_f^2-V_i^2}{2a}=\frac{0m/s^2-(13m/s)^2)}{2(-9.8m/s^2)}=8.62m\]
Add that to its initial height to get a total height of 98.62m.

Now calculate how long it takes to reach the ground:
\[d=\frac{1}{2}gt^2=\frac{1}{2}(-9.8m/s^2)t^2=98.62m\]\[t=4.486s\]
For the last part:\[V_f=V_i+at=0m/s+(-9.8m/s^2)(4.49s)=43.96m/s\]
You can also use conservation of energy to do the last part since all of the PE will be converted to KE on the way down:
\[PE=KE\]\[\cancel m gh=\frac{1}{2} \cancel m v^2\]\[(9.8m/s^2)(98.62m)=966.48J\]\[966.48J=\frac{1}{2}v^2\]\[v=\text{~}43.96m/s\]

Answer 17

@Shane_B u r right...i dint think abt it...thnxx

Answer 18

np :)