Projectile Motion balloon question

In summary, an expert threw a 1.0kg stone from a descending balloon and it hit the ground 6.20 seconds after being thrown. The stone's initial velocity was 14.8m/s perpendicular to the path of the descending balloon and its final position was 0m. The y velocity relative to the person in the basket when the rock was thrown was negative and the y acceleration relative to the basket was -60.7m/s^2.
  • #1
Elvis 123456789
158
6

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
 
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Likes jongo said
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  • #2
It says 'perpendicular to the path of the descending balloon'. What direction does that make it?
 
  • #3
oh wow ok. The initial Y velocity is negative. I got 334m
 
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Likes jongo said
  • #4
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
 
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Likes jongo said
  • #5
Elvis 123456789 said:
oh wow ok. The initial Y velocity is negative. I got 334m
yes, that looks about right.
 
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Likes jongo said
  • #6
Elvis 123456789 said:
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?
 
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Likes jongo said
  • #7
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?
 
  • #8
Elvis 123456789 said:
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?
 
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Likes jongo said
  • #9
That's what I calculated for the Y displacement of the stone when it reached the ground, relative to the balloon.
 
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Likes jongo said
  • #10
Elvis 123456789 said:
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.
 
  • #11
Thank you, and your right that would've been a much easier way. I tend to over complicate things more than I should.
 
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Likes jongo said

1. What is projectile motion?

Projectile motion is the motion of an object through the air, characterized by a curved path due to the influence of gravity.

2. How does a balloon demonstrate projectile motion?

When a balloon is released, it moves through the air in a curved path due to the force of gravity acting on it. This demonstrates projectile motion.

3. What factors affect the trajectory of a balloon in projectile motion?

The initial velocity of the balloon, the angle at which it is released, and air resistance are all factors that affect the trajectory of a balloon in projectile motion.

4. Why does the balloon eventually fall to the ground in projectile motion?

The balloon eventually falls to the ground due to the force of gravity acting on it. As the balloon moves through the air, the force of gravity pulls it towards the ground, causing it to eventually fall.

5. How is the motion of a balloon in projectile motion similar to that of other objects?

The motion of a balloon in projectile motion is similar to that of other objects because it follows the same laws of motion, including Newton's laws of motion and the law of gravity. The only difference is the initial force that propels the balloon, which is typically provided by the person releasing it.

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