# Projectile motion -- throwing an object vertically

• hescot2
In summary, a man is standing at a launch site. The rocket doesn't liftoff. Frustrated, the man throws it straight up with a speed of 12.42 m/s. It is caught on the way down at a point 5.0 meters above where it was thrown, by a fellow rocketeer on the second floor, who doesn't want to see any harm come to the rocket. The first man wants to figure out how fast the rocket was going when it was caught. Again, thanks for all your help, really rusty with this right now.
hescot2
Homework Statement
.
Relevant Equations
.
I don't need to know how to solve this but I was just looking for the answer. Can anyone help?

A man is standing at a launch site. The rocket doesn't liftoff. Frustrated s/he throws it straight up with a speed of 12.42 m/s. It is caught on the way down at a point 5.0 meters above where it was thrown, by a fellow rocketeer on the second floor, who doesn't want to see any harm come ot this wonderful rocket. The first man wants to figure out how fast the rocket was going when it was caught. Again thanks for all your help, really rusty with this right now.

hescot2 said:
Homework Statement:: .
Relevant Equations:: .

I don't need to know how to solve this but I was just looking for the answer.
So it isn't homework?

no. I am too old for homework. just trying to solve with friends for a game

Is it ok to ask here?

hescot2 said:
Homework Statement:: .
Relevant Equations:: .

I don't need to know how to solve this but I was just looking for the answer. Can anyone help?

A man is standing at a launch site. The rocket doesn't liftoff. Frustrated s/he throws it straight up with a speed of 12.42 m/s. It is caught on the way down at a point 5.0 meters above where it was thrown, by a fellow rocketeer on the second floor, who doesn't want to see any harm come ot this wonderful rocket. The first man wants to figure out how fast the rocket was going when it was caught. Again thanks for all your help, really rusty with this right now.
hescot2 said:
no. I am too old for homework. just trying to solve with friends for a game

Is it ok to ask here?

Welcome to the PF. No, it is against the rules to post schoolwork type questions and not show any work. That's not how the PF works. Here is a link that helps to explain the reasons for us treating all homework and schoolwork-type threads this way:

Hope that makes sense. You will probably still find the PF useful as a learning tool, but we won't solve problems like this for you. We can give you a couple hints and point you to the equations that you will need to understand and use to solve this problem, though. Let us know if you want the hints and pointers to the equations to use.

Sorry! Are they equations that a 60 year old could do?? I am so far from doing anything like this. I appreciate any help that could streer me in the right direction.

One of the equations you can use when you're dealing with constant acceleration is$${v_y}^2 = {u_y}^2 + 2a_y \Delta y$$where ##u_y = 12.42 \text{ms}^{-1}##, ##a_y = -g##, ##\Delta y = 5.0 \text{m}##. Hopefully you see that there are two solutions for ##v_y## of equal magnitude, and you're interested in the solution at time ##t_2##, with ##t_2 > t_1##, where ##v_y < 0##.

berkeman
And ##g## it the acceleration due to gravity, which is ##9.8\frac{m}{s^s}## for this problem.

Try working on that equation and post your work, and we can try to help you through it. Think of it as a learning adventure...

etotheipi
I've got almost a decade on you, so yes we do. Did you take physics in school?
I see @etotheipi has used conservation of energy to supply you a good start.

etotheipi
thank you for the response but I understand none of that. I guess that I am just too old. sorry for bothering you and thank you for helping as much as you were allowed to! If you are ever interested in the game that this applies to,

https://www.geocaching.com/geocache/GCKNKX

someone in our group came up with 9.365. Are you allowed to verify that?

hescot2 said:
someone in our group came up with 9.365. Are you allowed to verify that?

Sorry, it's wrong.

ugg. Thanks.

hescot2 said:
ugg. Thanks.
Maybe you could get whoever came up with that number to explain the reasoning and post that. Then we can progress.

hutchphd

hescot2 said:
Yes.

THANK YOU!

hutchphd
Do we win a prize?

etotheipi said:
One of the equations you can use when you're dealing with constant acceleration is$${v_y}^2 = {u_y}^2 + 2a_y \Delta y$$
Personally, that is the one SUVAT equation that I have never bothered to memorize. But divide by two and multiply by m and you get the equation that I use all the time:$$\frac{1}{2}mv_y^2 = \frac{1}{2}mu_y^2 + ma_y\Delta y$$Which of course is an assertion about kinetic energy after an acceleration in terms of kinetic energy before the acceleration plus work done by the acceleration.

etotheipi
Sure that works! You can also get it with $$\Delta y = \frac{1}{2}(u_y + v_y)t = \frac{1}{2} (u_y + v_y) \frac{(v_y - u_y)}{a_y} = \frac{v_y^2 - u_y^2}{2a_y}$$

PeroK

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air that is subject to only the force of gravity. It is a type of motion that follows a curved path, known as a parabolic path, due to the combination of the object's initial velocity and the constant force of gravity acting on it.

## 2. How does throwing an object vertically affect its motion?

Throwing an object vertically means that the initial velocity of the object is in the vertical direction only. This causes the object to move upward, reach a maximum height, and then fall back down due to the force of gravity. The object's motion follows a parabolic path, with the highest point being the maximum height.

## 3. What factors affect the trajectory of a vertically thrown object?

The trajectory of a vertically thrown object is affected by the initial velocity, the angle at which the object is thrown, and the force of gravity. The greater the initial velocity and the smaller the angle of throw, the higher the object will go. The force of gravity always acts in the downward direction, causing the object to fall back to the ground.

## 4. How can we calculate the maximum height and time of flight for a vertically thrown object?

The maximum height of a vertically thrown object can be calculated using the formula h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the angle of throw, and g is the acceleration due to gravity. The time of flight can be calculated using the formula t = 2v sinθ/g.

## 5. Is the motion of a vertically thrown object affected by air resistance?

Yes, air resistance can affect the motion of a vertically thrown object by slowing it down as it moves through the air. However, for most objects thrown at typical speeds, the effect of air resistance is negligible and can be ignored in calculations. This is why projectile motion is often studied in a vacuum or in idealized conditions without air resistance.

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