# Kinematics: Free fall problem.

• NoahCygnus
In summary, the homework statement is saying that a ball is thrown vertically upward with a speed 'v' from a height 'h' metre above the ground. The time taken for the ball to hit the ground is found by solving a quadratic equation. The equation used is s = ut + \frac{1}{2}at^2. The attempt at a solution was to solve for t1 using the equation y(t) = h + vt - \frac{1}{2}gt^2. The results of the attempt did not match the answer given in the book, so t was found by solving the original equation with s = -h. Finally, the thread was left here.
NoahCygnus

## Homework Statement

"A ball is thrown vertically upward with a speed 'v' from a height 'h' metre above the ground. The time taken for the ball to hit the ground is."

## Homework Equations

##s = ut + \frac{1}{2}at^2##

## The Attempt at a Solution

https://scontent.fdel3-1.fna.fbcdn.net/v/t1.0-9/17352002_706036796237108_6058839347900308786_n.jpg?oh=3f0a769b256f51becc5ee351dc25294b&oe=596F1319
https://scontent.fdel3-1.fna.fbcdn.net/v/t1.0-9/17264650_706036792903775_4158276045728207185_n.jpg?oh=ab150cc42047d1d745b8b84a53fe094e&oe=596262E0
So we throw the ball vertically upward with a velocity 'v' from a reference point 'o' as I have labelled in the diagram. The ball returns back to the reference point after an interval of t1, and the displacement in this time period is 0. I used the equation ##s = ut + \frac{1}{2}at^2## and solved the quadratic equation for t1.

Now the ball is back at o, but this time it has velocity -v, and is falling down. I assumed it took the ball time t2 to fall from o to h. Again using the equation ##s = ut + \frac{1}{2}at^2## and solving the quadratic equation for t2, I calculated t2.

Then I added t1 and t2 for the total time taken for the fall to reach the ground. But my answer doesn't match the answer given in book. The answer given in the book is ##t = \frac{v}{g}[1+\sqrt{1+\frac{2gh}{v^2}}]## . I wonder what did I get wrong. Thank you for your help.

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NoahCygnus said:
I wonder what did I get wrong
I wonder too. Show your steps, so we can try to pinpoint it...

You can work it out in one go without having to analyze two parts of the motion. Think about it. For example, define the ground as the origin for the position, so now you have an initial position.
$$y(t) = h + vt - \frac{1}{2}gt^2$$
Do you know what you need to do now to solve for the required time?

NoahCygnus
Although: do you think the results are very different ?

NoahCygnus
And: you want to get rid of the ##\pm## sign: there is only one answer...

BvU said:
I wonder too. Show your steps, so we can try to pinpoint it...
You can check my steps, I have uploaded photographs.
BvU said:
Although: do you think the results are very different ?
I am such an idiot, I forgot to manipulate my answer to match the answer given in the book when my answer does indeed match with the one given in book. I apologise, for the unnecessary bother. Thank you.
Anyway, how do I know that I have to remove ##\pm## sign?

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You could also use your original equation with s = -h to solve for t

NoahCygnus
NoahCygnus said:
You can check my steps i

I am such an idiot, I forgot to manipulate my answer to match the answer given in the book when my answer does indeed match with the one given in book. I apologise, for the unnecessary bother. Thank you.
Anyway, how do I know that I have to remove ##\pm## sign?
You need to see which makes sense physically. If a specific sign ends up giving you a negative answer, when time can't be negative (initial value problems like this one mean the time starts at 0 and cannot be anything less), you can discard that sign.

JoePhysics said:
You need to see which makes sense physically. If a specific sign ends up giving you a negative answer, when time can't be negative (initial value problems like this one mean the time starts at 0 and cannot be anything less), you can discard that sign.
I see. Thanks. Can I remove this thread, or is it a policy of physics forums to keep all threads, no matter how idiotic they are? (Like mine, as I already had the answer).

NoahCygnus said:
I see. Thanks. Can I remove this thread, or is it a policy of physics forums to keep all threads, no matter how idiotic they are? (Like mine, as I already had the answer).
Leave it here. fwiw, your thread was not idiotic at all, no question ever is. Look at it this way, andrevdh and me also gave you a quicker way to arrive at your answer, which may prove useful to you in the future, and you also learned what to check to see when you can remove a sign from the solution of a quadratic.

NoahCygnus
andrevdh said:
You could also use your original equation with s = -h to solve for t
That surely reduces all the work I had to do to solve the problem. I didn't know the original equation was powerful enough to give me answer in one go. I am new to physics, so I don't have much experience. I couldn't imagine in my head how the above equation could account for the time interval of the whole motion.

NoahCygnus said:
That surely reduces all the work I had to do to solve the problem. I didn't know the original equation was powerful enough to give me answer in one go. I am new to physics, so I don't have much experience. I couldn't imagine in my head how the above equation could account for the time interval of the whole motion.
In essence, as long as the ball is subject exclusively to the uniform acceleration of gravity, the equations apply. The moment it hits the ground, or if there's a gust of wind, or if anything changes the initial hypotheses (ball subject exclusively to uniform gravity) then the equation is no longer valid. In your example, from the initial moment right up to when it hits the ground, the equation holds and it describes the motion of the ball.

Don't worry I also did not realize it when I started out, but it helps if you do not just jump in. Try to think if there could be other ways to solve the problem.

## 1. What is free fall?

Free fall is the motion of an object under the influence of only gravity, with no other forces acting on it. This means that the object is accelerating towards the ground at a constant rate of 9.8 meters per second squared.

## 2. How is free fall different from other types of motion?

Free fall is different from other types of motion because it is a special case of motion where the only force acting on the object is gravity. This means that the acceleration of the object is constant and independent of its mass or any other external factors.

## 3. How do you calculate the distance traveled in free fall?

The distance traveled in free fall can be calculated using the formula d = 1/2 * g * t^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time the object has been falling. This formula assumes that the object was dropped from rest.

## 4. What is the equation for velocity in free fall?

The equation for velocity in free fall is v = g * t, where v is the velocity and t is the time the object has been falling. This means that the velocity of an object in free fall increases by 9.8 meters per second for every second it falls.

## 5. How can I use kinematics to solve free fall problems?

To solve free fall problems using kinematics, you will need to use the three kinematic equations: d = v0t + 1/2at^2, v = v0 + at, and v^2 = v0^2 + 2ad. These equations can be used to find the distance, velocity, and acceleration of an object in free fall, given any two of these variables.

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