# Kinematics: Catch Ball Dropped from Balloon After 24.7s

• negation
In summary, the conversation discussed the problem of a balloon rising at a constant velocity while a passenger throws a ball straight up at a different velocity. The goal was to determine how much time would pass before the passenger catches the ball again. The conversation covered different equations and approaches to solving the problem, ultimately concluding that the ball and balloon would meet again in 2.4 seconds.
negation

## Homework Statement

A balloon is rising at 10m/s when its passenger throws a ball straight up at 12m/s relative to the balloon . How much later does the passenger catches the ball?

## The Attempt at a Solution

vf^2 - vi^2 = 2g(yf-yi)

Ball:
0ms^-1 - (22ms^-1)^2 = 2(-9.8ms^-2)(yball)
yball = 24.7m

Balloon:
0ms^-1 - (10ms^-1)^2 = 2(-9.8ms^-2)(yballoon)
yballoon = 24.7m

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Assume that the velocity of the balloon is constant.

voko said:
Assume that the velocity of the balloon is constant.
You're right since the ball is thrown up relative to the balloon.

Last edited:
The balloon is not a projectile, there are other forces acting on it. That does not mean, in general, that the acceleration is zero, but since you are not given any, the only sane assumption is that the acceleration is zero.

yf - yi = vit + 0.5gt^2

Ball:

yball = (12ms^-1 + 10ms^-1)t + 0.5(-9.8ms^-2)t^2

Balloon:

yballoon = 10ms^-1 t + 0.5(oms^-2)t^2
y balloon = 10ms^-1

yballoon = yball

10ms^-1 t + 0.5(oms^-2)t^2 = 10ms^-1
t = 2.4s

There is a simpler approach. Since the balloon is not accelerated, everything can be computed with regard to the balloon as if it was stationary. So that basically means the projectile goes up at 12 m/s. At some time t it will reach a maximum elevation (from the balloon). This time can be easily computed from the condition that the (relative) velocity becomes zero. Going down takes the same time.

voko said:
There is a simpler approach. Since the balloon is not accelerated, everything can be computed with regard to the balloon as if it was stationary. So that basically means the projectile goes up at 12 m/s. At some time t it will reach a maximum elevation (from the balloon). This time can be easily computed from the condition that the (relative) velocity becomes zero. Going down takes the same time.

So is 2.4s the right answer?
It appears more logical if I multiply 2.4s by 2.

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negation said:
So is 2.4s the right answer?
It appears more logical if I multiply 2.4s by 2.
You correctly calculated 2.4s as the time when ball and balloon would meet again. If you had calculated the time to when the relative velocity were zero, then you would have got 1.2s and would need to double for the second half.
In fact, a slightly simpler calculation than the one you did is to say that they will meet again when the relative velocity is equal and opposite to its initial value, giving gt = 24m/s.

## What is kinematics?

Kinematics is the branch of classical mechanics that describes the motion of objects without considering the causes of motion.

## What is the significance of catching a ball dropped from a balloon after 24.7 seconds?

This scenario is commonly used in physics experiments to study the effects of gravity on an object's motion. By measuring the time it takes for the ball to fall and the distance it travels, we can calculate the acceleration due to gravity.

## What factors affect the motion of the ball in this scenario?

The main factors that affect the motion of the ball are gravity, air resistance, and the initial velocity of the ball when it is dropped from the balloon. The height from which the ball is dropped can also play a role in the ball's motion.

## How can we calculate the acceleration due to gravity from this experiment?

Using the formula a = (2d)/t^2, where a is the acceleration due to gravity, d is the distance the ball falls, and t is the time it takes to fall, we can determine the acceleration due to gravity.

## What other real-life applications does kinematics have?

Kinematics is used in various fields such as engineering, sports, and animation to analyze and predict the motion of objects. It is also used in navigation systems, robotics, and space exploration to calculate trajectories and movements of objects in motion.

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