# Constant acceleration.

## 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.

Doc Al
Mentor
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?)

## 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
Mentor
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.

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
Mentor
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.

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
Mentor
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).