Projectile Motion balloon question

[Elvis 123456789]

Homework Statement

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 23.5m/s A 1.0-kg stone is thrown from the basket with an initial velocity of 14.8m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.20s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 23.5m/s.

How high was the balloon when the rock was thrown out?

Homework Equations

ΔY=V1y*t + 1/2*a*t^2

The Attempt at a Solution

At first I thought since the stone is being thrown off the balloon which is going at a constant speed of 23.5m/s, then the stone's initial Y velocity must also be 23.5m/s, and since its being thrown perpendicular to the balloon's path at 14.8m/s, then that must be its initial X velocity.

I figure that the balloon's height when the stone was thrown out, is the same as the stone's initial position. And the stone's final position is 0 at t=6.20 s since its on the ground.

So I used the above equation
ΔY=V1y*t + 1/2*a*t^2

0-y1=V1y*t - 1/2*g*t^2

y1= -[(23.5m/s)(6.20s)-(4.9m/s^2)(6.20s)^2
y1=47.356m

but this isn't right and I don't know why

 

[haruspex]

It says 'perpendicular to the path of the descending balloon'. What direction does that make it?

 

[Elvis 123456789]

oh wow ok. The initial Y velocity is negative. I got 334m

 

[Elvis 123456789]

Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket.

I'm having trouble figuring out the above part. I know the X velocity is constant, but i don't know how to get the Y velocity. I tried using the equation V^2=V1^2 + 2aΔY

V^2=(-23.5m/s)^2 -2*(9.8m/s^2)*188m

and i know this is wrong so idk

 

[haruspex]

oh wow ok. The initial Y velocity is negative. I got 334m

yes, that looks about right.

 

[haruspex]

Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer at rest in the basket.

I'm having trouble figuring out the above part. I know the X velocity is constant, but i don't know how to get the Y velocity. I tried using the equation V^2=V1^2 + 2aΔY

V^2=(-23.5m/s)^2 -2*(9.8m/s^2)*188m

and i know this is wrong so idk

What was the y velocity relative to the person in the basket when the rock was thrown? What was the y acceleration relative to the basket?

 

[Elvis 123456789]

would it be

V^2=-2*(9.8m/s^2)*(-188m)
V=60.7 but it would be negative since the velocity is in the negative direction?

 

[haruspex]

would it be

V^2=-2*(9.8m/s^2)*(-188m)
V=60.7 but it would be negative since the velocity is in the negative direction?

Where are you getting the 188m from?

 

[Elvis 123456789]

That's what I calculated for the Y displacement of the stone when it reached the ground, relative to the balloon.

 

[haruspex]

That's what I calculated for the Y displacement of the stone when it reached the ground, relative to the balloon.

Ok. Yes, that's the right answer. Your method produces an ambiguity in the sign. You can avoid that (and the need to take a square root) by using the time and relative acceleration.

 

[Elvis 123456789]

Thank you, and your right that would've been a much easier way. I tend to over complicate things more than I should.