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Constant acceleration.

  • Thread starter brusier
  • Start date
  • #1
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Homework Statement


A hot air balloon ascending at a rate of 10m/s drops a package when the balloon reaches a height of 73 m. Find time to ground. Find velocity when package hits ground.


The balloon and package, in this case, can be treated as a point particles and as such can be manipluated with genral kinematics equations.

I apologize for my inability to display mathematical symbols when creating equations....


Homework Equations





X = X(initial) + V(initial)t + 1/2at^2

V(final)^2 = V(initial)^2 + at


The Attempt at a Solution




I simply used the quadratic formula with the first equation to solve for t because the equation is already in terms of t. This produced:

-10(+-)sq.rt(100+1460) / -10

where:
X=0
X(initial)=73
V(initial)= 10m/s
a= -10m/s


I chose the negative value of b^2 - 4ac to produce:

t = -49.49/-10 = 4.949s

Secondly, I used this time in the V(final) = V(initial) + at equation:

V(final) = 10 + (-10*4.949) = -39.49m/s


Am I thinking about this scenario correctly enough so that I am using the appropriate equations?? The answers, I am told, are incorrect.
 

Answers and Replies

  • #2
Doc Al
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I don't see anything wrong with your method or answers. Why do you think they are incorrect? (Are you allowed to use g = 10, instead of 9.8 m/s^2?)
 
  • #3

Homework Equations




V(final)^2 = V(initial)^2 + at
sorry if i'm being stupid, but i think this equation is wrong.
shouldn't it ought to be
V(final) = V(initial) + at?


I get the same t and V(final), so yeah. it doesn't matter.
 
  • #4
Doc Al
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44,882
1,129
sorry if i'm being stupid, but i think this equation is wrong.
shouldn't it ought to be
V(final) = V(initial) + at?
You're correct: The original post did list an incorrect equation. But the correct version was used to solve the problem.
 
  • #5
27
0
I am extremely sorry for the error!!

In laboriously typing the equation out I inadvertently mixed the

V(final)^2 = V(initial)^2 + 2ax equation with the V(final) = V(initial) +at equation.


very sorry!! I see that I also got my units wrong for a or, in this case, g! ouch.

thanks for the help.

I'm sure I am able to use constant g = -10m/s^2.

The homework is posted online through a webassign account and since it is my first experience using it, I hope my unfamiliarity is causing the confusion, not my inaccurate computations! lol
 
  • #6
Doc Al
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44,882
1,129
I am extremely sorry for the error!!

In laboriously typing the equation out I inadvertently mixed the

V(final)^2 = V(initial)^2 + 2ax equation
Since you are familiar with this kinematic equation, why not use it to check your answer for the final speed without using the time.
 
  • #7
27
0
Since you are familiar with this kinematic equation, why not use it to check your answer for the final speed without using the time.
V(final)^2 = V(initial)^2 + 2ax:

Vf^2 = 10^2 + (2)(-10)(-73)

Vf^2 = 100 + 1460

Vf^2 = 1560

Vf = 39.49 m/s

should this vector quantity be made negative to indicate downward motion?
 
  • #8
Doc Al
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should this vector quantity be made negative to indicate downward motion?
Yes. Or you can just specify the direction as downward.

Whenever you take a square root you get two answers: one +, one -. It's up to you to choose the one that fits the situation. In this case, you know it's moving downward, thus you'd choose the negative answer (using the usual sign convention).
 

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