A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.0 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground. After it is released, the sandbag is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.0 s after its release. (b) How many seconds after its release will the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches?

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A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.0 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground. After it is released, the sandbag is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.0 s after its release. (b) How many seconds after its release will the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches?

Physics
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will also try to answer this :D
|dw:1439963429493:dw|
At the instant the sand bag is dropped, its velocity will be 5 m/s upwards. Do you want to attempt it on your own first?

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maybe laterrr. I'm finishing my other problems :D
|dw:1439963731904:dw|
Since, I have to go to, i'll leave you some hints. To find the velocity at given times, you can use the equation v=u+gt Here, u will be 5m/s. To find the positions at given time you can use s = ut + 1/2gt^2 Don't forget to use the sign convention.
At greatest height velocity will become 0, you can use this fact to find the greatest height that the sand bag reaches.
Tag me if you get stuck anywhere c:
will do sir. Thank you!
Okay I'll start answering now!
t at 0.250 Y=(5m/s)(.250s)-1/2(9.8)(.250s) = 0.94375 m @Abhisar am I on the right track?
t at 1 s Y = 0.451 m ?
t at .250 Vf = 5m/s - (9.8)(.250) Vf = 2.55 t at 1 Vf = -4.8 m/s
am I doing it right looool
b.) is quite confusing hmmm doI have to use 2 equations???
Your first answer is correct, let's see what time does the sand bag takes to reach the maximum height. At max height V=0, putting this value in the equation v=u+at we get that t= 0.5. So in 1 sec the bag will return to its initial position i.e. 40m from the ground. Getting it?
Exact position at 0.25 seconds will be 40+0.94 m from ground.
hmmm okay. Wait I'll try to use your concept
where did you t = 0.5??
where did u get* @Abhisar
Using 1st law of motion, V=U+gt, Since, at the maximum height velocity of the bag will become 0, V=0. Also u=5m/s => 0=5-10t (taking g=10m/s^2) =>t=0.5 sec
For the part b you have to calculate time in 2 steps as following: Step 1: since initially bag will move upward for some time (0.5 sec) and then come back to its initial position (40 m from ground) again taking time equal to its time of ascend (0.5s). So total time taken in the cycle (going up and then coming down) will be 0.5+0.5 = 1s. Step 2: Now we need to find how much time the bag will take to reach the ground from 40 m above it. We can use s=ut+at^2. (u will again be 5m/s but in downward direction this time) Total time taken will be the sum of time calculated in step 1 and step 2
so you mean to say 40=(-5)(t)+1/2(9.8)(t)^2 ???
I got 3.4125 s
Total time will be \(\approx 3.38~s\)

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