anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
will also try to answer this :D
Abhisar
  • Abhisar
|dw:1439963429493:dw|
Abhisar
  • Abhisar
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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anonymous
  • anonymous
maybe laterrr. I'm finishing my other problems :D
Abhisar
  • Abhisar
|dw:1439963731904:dw|
Abhisar
  • Abhisar
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.
Abhisar
  • Abhisar
At greatest height velocity will become 0, you can use this fact to find the greatest height that the sand bag reaches.
Abhisar
  • Abhisar
Tag me if you get stuck anywhere c:
anonymous
  • anonymous
will do sir. Thank you!
anonymous
  • anonymous
Okay I'll start answering now!
anonymous
  • anonymous
t at 0.250 Y=(5m/s)(.250s)-1/2(9.8)(.250s) = 0.94375 m @Abhisar am I on the right track?
anonymous
  • anonymous
t at 1 s Y = 0.451 m ?
anonymous
  • anonymous
t at .250 Vf = 5m/s - (9.8)(.250) Vf = 2.55 t at 1 Vf = -4.8 m/s
anonymous
  • anonymous
am I doing it right looool
anonymous
  • anonymous
b.) is quite confusing hmmm doI have to use 2 equations???
Abhisar
  • Abhisar
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?
Abhisar
  • Abhisar
Exact position at 0.25 seconds will be 40+0.94 m from ground.
anonymous
  • anonymous
hmmm okay. Wait I'll try to use your concept
anonymous
  • anonymous
where did you t = 0.5??
anonymous
  • anonymous
where did u get* @Abhisar
Abhisar
  • 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
Abhisar
  • Abhisar
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
anonymous
  • anonymous
so you mean to say 40=(-5)(t)+1/2(9.8)(t)^2 ???
anonymous
  • anonymous
I got 3.4125 s
Abhisar
  • Abhisar
Total time will be \(\approx 3.38~s\)

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