Acceleration and Gravity: Two Problems to Solve Using Kinematic Equations

[Mike_Winegar]

Hey everyone, this is my first time using the PF, so please excuse me if I am not following the usual method for posting questions. Also, its my first time using Latex spelling, so i may very well have slaughtered it.

I have access to 5 equations...

1.$$v=v_0+a t$$
2.$$\Delta X=v_0 t+/frac{1} {2} a t^2$$
3.$$v^2=v_0^2 +2a delta x$$
4.$$\Delta X=/frac{1}{2}(v_0 + v)t$$
5.$$\Delta X=v t - /frac{1} {2}a t^2$$

My teacher has given us 2 problems, and the only help he gave was "use the formulas!".

The problems are as follows...

1. A rifle is fired and the bullet is accelerated from rest through the barrel which is 1.00m long. If the bullet leaves the barrel at a speed of 600m/s, calculate its acceleration.

2. A balloon is rising at a steady rate of 29.4 m/s. A stone falls from the balloon and reaches the ground in 20.0 seconds. At the instant the stone reaches the ground, how far is the balloon above it?

I have thought about both of the equations and I have come up with the following...

1. I really have no idea on this one...The only idea I have been able to come up with is if it leaves the 1m long barrel at a speed of 600m/s then it took 1/600th of a second to travel the length of the barrel.

2. I don't know which formula to use on this one, but I did some different kind of work on it. If the balloon is rising at 29.4 m/s and it takes the stone 20 seconds to fall to the ground, wouldn't I just be able to say 29.4 m/s * 20?

Help on these would be extremely appreciative. :!)

 

[Mike_Winegar]

Jeeze...It seems that I can't get the Latex spelling right...what the formulas are supposed to be are...

2. Delta X = v_0t + .5at^2
3. v^2=v_0^2+ 2a*Delta X
4. Delta X = .5(v_0+v)t
5. Delta X = vt- 1/2 at^2

 

[Max Eilerson]

1. You know the initial velocity, final velocity and the distance travelled. You want to find the acceleration a, which equation has all four of these variables. Just rearrange it to find a and put the numbers in.

 

[Max Eilerson]

2. I don't know which formula to use on this one, but I did some different kind of work on it. If the balloon is rising at 29.4 m/s and it takes the stone 20 seconds to fall to the ground, wouldn't I just be able to say 29.4 m/s * 20?
)

That would be how far the balloon has traveled from the point where the stone was released. You need to find the distance the stone fell and then add these two together, think about the initial velocity of the stone when it's released and then find the relevant formula for x.

 

[Mike_Winegar]

I did some more work and I got 180,000 m/s^2 for number 1. Is this correct?

 

[Max Eilerson]

I did some more work and I got 180,000 m/s^2 for number 1. Is this correct?

Yes. No. 2 shouldn't be hard.

 

[Mike_Winegar]

Thanks! I'm currently working on number 2...stick around please :)

 

[Mike_Winegar]

Alright, I got 1960 total meters between the rock and the balloon. Is this correct?

 

[Mike_Winegar]

I tried it again, and I got 1915.9 =/ Am I even close?

 

[Max Eilerson]

1960m is correct. Post your working.

 

[Mike_Winegar]

For the first 3 seconds, the rock has the initial velocity 29.4m/s, the final velocity is 0. The acceleration is -9.8m/s^2. Plug that into the formula, and you end up with 44.1m traveled by the rock until its velocity is = 0?

For the following 17 seconds, the rock has the initial velocity 0, the final velocity is 166.6m/s. The acceleration is 9.8m/s^2. Plug that into the formula, and you get 1416.1m, add 588m for the distance the balloon traveled, and you get 2004.1, subtract 44.1 for the distance the rock traveled upwards and you get 1960m total distance between the balloon and the rock.

Does this look correct?

 

[Max Eilerson]

Yes, and all that is equal to 0.5at^2.

 

[Mike_Winegar]

Wow, Thanks so much for all your help! I'll be sure to come back again! :)