Finding Time and Velocity for a Package Dropped from a Hot Air Balloon

[brusier]

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 general 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.

 

[Doc Al]

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?)

 

[earlofwessex]

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.

 

[Doc Al]

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.

 

[brusier]

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

 

[Doc Al]

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.

 

[brusier]

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?

 

[Doc Al]

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).