# How Long Does It Take for a Dropped Ball to Hit the Ground?

• Govind_Balaji
In summary: We can invert the matrix and get:$$\begin{pmatrix} x\\ y\end{pmatrix}=\begin{pmatrix} \frac{d}{ad-bc} & \frac{-b}{ad-bc}\\ \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{pmatrix}\begin{pmatrix} ...\\ ...\end{pmatrix}$$...but it is usually more efficient to use Cramer's rule if you need to solve simultaneous equations - even if you do not know how to find determinants.In summary, a ball is dropped from a balloon going up
Govind_Balaji

## Homework Statement

A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take to reach the ground?

Ans:4.3 seconds

## The Attempt at a Solution

I can't understand the question.
Is the initial velocity of the ball when it is dropped=0? or is it 7m/s.

What value should I take for g? Should I take g= 9.8 m/s$^2$?
Please help me. I assumed ball is dropped from 60 m height with initial velocity=0 and a=g=9.8 m/s$^2$. But it gives wrong answer. What is the concept of this question? How to solve it?

Govind_Balaji said:

## Homework Statement

A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take to reach the ground?

Ans:4.3 seconds

## The Attempt at a Solution

I can't understand the question.
Is the initial velocity of the ball when it is dropped=0? or is it 7m/s.

What value should I take for g? Should I take g= 9.8 m/s$^2$?
Please help me. I assumed ball is dropped from 60 m height with initial velocity=0 and a=g=9.8 m/s$^2$. But it gives wrong answer. What is the concept of this question? How to solve it?

g = 9.8 m/s2 is an acceptable value for g unless otherwise instructed.

The ball is being dropped (which means simply released) from a moving platform. It shares the motion of the platform until the instant it is released after which its trajectory is independent ("free fall").

1 person
Let v be the final velocity when it reaches the ground.
Let x be the distance=60 m
Let u be the initial velocity=0 m/s

a=g=9.8 m/s

By 2nd equation of motion,
$x=ut+\frac{at^2}{2}$
$x=0*t+\frac{9.8*t^2}{2}$
$60=\frac{9.8*t^2}{2}$
$120=9.8t^2$
$t^2=\frac{120}{9.8}$
$t^2=2.44$
$t=\sqrt{2.44}$
$t=1.56$

Actually I should get $t=4.3$

Is the initial velocity of the ball when it is dropped=0? or is it 7m/s.
What arguments from the problem statement can you make for either possibility? Something to do with the velocity of the balloon?

Also remember that velocity is a vector.

What value should I take for g? Should I take g= 9.8 m/s2
Acceleration is also a vector - is the acceleration of the ball in the same direction as the initial velocity?

If you make "downwards" positive - what are the values for initial velocity and acceleration going to be.

Reconsider the initial velocity of the ball.

1 person
gneill said:
Reconsider the initial velocity of the ball.

I wrote everything that was in my textbook. I said earlier that I didn't understand the question. I assumed the initial velocity to be 0. I am not certain.

Simon Bridge said:
What arguments from the problem statement can you make for either possibility? Something to do with the velocity of the balloon?

Also remember that velocity is a vector.

Acceleration is also a vector - is the acceleration of the ball in the same direction as the initial velocity?

If you make "downwards" positive - what are the values for initial velocity and acceleration going to be.
Yes I made downwards positive since it was all about downwards motion.

My friend said that momentum of the balloon slows down the velocity of the ball. I think it is not right. Am I correct?

I can see you are struggling to understand the problem - but if you do not explain how you are thinking about it, we can only guess what you are having trouble with. Don't worry about sounding silly - we've all done that: we understand.

OK - imagine someone in the balloon holding the ball out over the side.
The balloon is going upwards, before the ball is released: what is the ball doing?

At the instant the ball is released, it still has the same velocity as just before it was released.
Now do you see?

If "downwards" is positive, what is the sign of the initial velocity of the ball?

1 person
gneill said:
Reconsider the initial velocity of the ball.

Simon Bridge said:
What arguments from the problem statement can you make for either possibility? Something to do with the velocity of the balloon?

Also remember that velocity is a vector.

Acceleration is also a vector - is the acceleration of the ball in the same direction as the initial velocity?

If you make "downwards" positive - what are the values for initial velocity and acceleration going to be.

Thank you both, I found that initial velocity=-7 m/s. I also learned that in a moving platform, an object in the platform shares the velocity with the platform. I found this by comparing me traveling in a train.

Here's my new attempt:

$$u=-7 m/s$$

##x=60 m##

##g=9.8 m/s^2##

##x=ut+\frac{at^2}{2}##
##60=-7t+\frac{(9.8)t^2}{2}##
##60=\frac{-14t+9.8t^2}{2}##
##120=-14t+9.8t^2##
Using quadratic equation formula, I got t=4.28$\approx$4.3 s

Well done:
You'd have got +4.3s and -2.9s ... you want the time that is in the future.
When you do long answers you should show that step.

BTW: good LaTeX use.

Reality check:
This time should be longer than if the ball was just dropped from a stationary balloon.
The calculation neglects air resistance. IRL air resistance cannot be neglected for such a long fall.

Simon Bridge said:
BTW: good LaTeX use.

Thank you. actually I started typing the answer before your last post. It took a long time to type LaTex(15-20 mins!)

Govind_Balaji said:
Thank you. actually I started typing the answer before your last post. It took a long time to type LaTex(15-20 mins!)
It gets faster with practice.

BTW the "itex" boxes are for inline use (where an equation sits inside a paragraph), while "tex" boxes are used for display equations - those that get their own line. The display form is best when you have lots of exponents or fractions.

i.e. a quadratic ##ax^2+bx+c=0## (1) can be solved by: $$x\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}\qquad \small{\text{...(2)}}$$... (1) is the "standard form" and (2) is the "quadratic equation".

If I put (2) in the inline form, it comes out like ##x\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}##

You can do multi-line equations... watch this:

\begin{align} & & ax+by=e & \qquad ...(1) \\ & & cx+dy=f & \qquad ...(2) \\ (1)\rightarrow & & y=\frac{e-ax}{b} & \qquad ...(3)\\ (3)\rightarrow (2) & & cx+d\frac{e-ax}{b}=f & \\ & & \implies x =\frac{bf-de}{bc-ad} & \qquad ...(4) \end{align}

Of course (1) and (2) can be written as a matrix equation:
$$\begin{pmatrix} a & b\\ c & d \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix} =\begin{pmatrix} e\\ f\end{pmatrix}$$
... just showing off :)

Last edited:

## 1. What is acceleration under gravity?

Acceleration under gravity is the rate at which an object's velocity increases when dropped or thrown near the Earth's surface. It is a constant value of approximately 9.8 meters per second squared (m/s^2) and is denoted by the letter 'g'.

## 2. How is acceleration under gravity measured?

Acceleration under gravity can be measured using a device called an accelerometer, which measures the change in velocity of an object over time. It can also be calculated using the formula a = g = Δv/Δt, where 'a' is the acceleration, 'g' is the acceleration due to gravity, and Δv/Δt is the change in velocity over a specific time period.

## 3. Does acceleration under gravity vary on different planets?

Yes, acceleration under gravity varies on different planets depending on their mass and size. The larger the planet, the higher the acceleration under gravity. For example, the acceleration under gravity on Mars is about 3.7 m/s^2, while on Jupiter it is about 24.8 m/s^2.

## 4. How does air resistance affect acceleration under gravity?

Air resistance, also known as drag, can affect acceleration under gravity by slowing down the rate at which an object falls. This is because air resistance creates an opposing force on the object, which reduces its acceleration. Objects with larger surface areas, such as parachutes, experience more air resistance and therefore have a slower acceleration under gravity.

## 5. What is the difference between acceleration under gravity and gravitational force?

Acceleration under gravity and gravitational force are related but distinct concepts. Acceleration under gravity is the acceleration an object experiences due to the Earth's gravitational force. Gravitational force, on the other hand, is the attractive force between two objects with mass. The value of gravitational force depends on the masses of the two objects, their distance apart, and the universal gravitational constant (G).

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