# Question about Projectile Motion: Throwing a Flower Pot from a Balcony

• cherryLoop
cherryLoop
Thread moved from the technical forums to the schoolwork forums
TL;DR Summary: I need help with solving a kinematics problem, thanks!

Brynn is standing on a balcony that is 40 m above the ground. She is surprised by a pigeon and throws a flower pot up in the air. It takes 3.5 s for the flowerpot to smash to the ground below. The flower pot experiences acceleration due to gravity of 9.8 m/s2[down]. Determine the initial velocity of the flower pot when Brynn released it and how fast the flowerpot was moving just before it smashed to the ground.

This question has been confusing me for a long time. I solved for the inital velocity by using delta d = v1xt + 1/2at^2, and using 3.5 seconds as time, acceleration as -9.8 and delta d as -40 meters. The answer I got was -5.7 m/s [up] but it confuses me because can the acceleration upwards be less than the acceleration downwards? And I'm not sure if my answer is correct. Could someone help correct me? Thank you so much.

cherryLoop said:
TL;DR Summary: I need help with solving a kinematics problem, thanks!

Brynn is standing on a balcony that is 40 m above the ground. She is surprised by a pigeon and throws a flower pot up in the air. It takes 3.5 s for the flowerpot to smash to the ground below. The flower pot experiences acceleration due to gravity of 9.8 m/s2[down]. Determine the initial velocity of the flower pot when Brynn released it and how fast the flowerpot was moving just before it smashed to the ground.

This question has been confusing me for a long time. I solved for the inital velocity by using delta d = v1xt + 1/2at^2, and using 3.5 seconds as time, acceleration as -9.8 and delta d as -40 meters. The answer I got was -5.7 m/s [up] but it confuses me because can the acceleration upwards be less than the acceleration downwards? And I'm not sure if my answer is correct. Could someone help correct me? Thank you so much.
I think you are confusing velocity with acceleration. There is only one acceleration throughout the motion of the flower pot and that it 9.8 m/s2 down.

If you need additional help, please post your work so that we can see what you did.

erobz
cherryLoop said:
acceleration as -9.8 and delta d as -40 meters. The answer I got was -5.7 m/s [up]
Why do you say that -5.7 m/s is "up"? You used negative to mean downwards for the other quantifies. Of course it is possible that the pot was thrown up initially, but you seem to be inconsistent in your sign convention.

A.T. said:
Why do you say that -5.7 m/s is "up"? You used negative to mean downwards for the other quantifies. Of course it is possible that the pot was thrown up initially, but you seem to be inconsistent in your sign convention.
It's in the statement of the problem.
cherryLoop said:
Brynn is standing on a balcony that is 40 m above the ground. She is surprised by a pigeon and throws a flower pot up in the air.

A.T. said:
Why do you say that -5.7 m/s is "up"? You used negative to mean downwards for the other quantifies.
I actually agree with you. It is indeed specified that the pot was thrown up, and that should give it an initial positive velocity, ending with a negative one by the time it hits the ground. Using a different sign between velocity and acceleration is going to result in mistakes.

kuruman said:
It's in the statement of the problem.
Indeed. Then the initial velocity should be positive.

Halc said:
I actually agree with you. It is indeed specified that the pot was thrown up, and that should give it an initial positive velocity, ending with a negative one by the time it hits the ground. Using a different sign between velocity and acceleration is going to result in mistakes.
I agree with all that. However, note the statement of the problem that instructs us to
cherryLoop said:
Determine the initial velocity of the flower pot when Brynn released it and how fast the flowerpot was moving just before it smashed to the ground.
"How fast the flowerpot was moving" indicates a speed not a velocity. For a example, a car's "speedometer" shows how fast the car is moving and has nothing to say about direction.

kuruman said:
"How fast the flowerpot was moving" indicates a speed not a velocity.
That should not be negative either.

Halc
A.T. said:
That should not be negative either.
I agree. Maybe one of the aims of the problem is to demonstrate that the initial and final speeds are the same but not the initial and final velocities. We need to see OP's work to find the error.

cherryLoop said:
TL;DR Summary: I need help with solving a kinematics problem, thanks!

Brynn is standing on a balcony that is 40 m above the ground. She is surprised by a pigeon and throws a flower pot up in the air. It takes 3.5 s for the flowerpot to smash to the ground below. The flower pot experiences acceleration due to gravity of 9.8 m/s2[down]. Determine the initial velocity of the flower pot when Brynn released it and how fast the flowerpot was moving just before it smashed to the ground.

This question has been confusing me for a long time. I solved for the initial velocity by using delta d = v1xt + 1/2at^2, and using 3.5 seconds as time, acceleration as -9.8 and delta d as -40 meters. The answer I got was -5.7 m/s [up] but it confuses me because can the acceleration upwards be less than the acceleration downwards? And I'm not sure if my answer is correct. Could someone help correct me? Thank you so much.
Plugging your values, ##\ a=-9.8\,,\ \ t=3.5\,,\ ## and ##\ \Delta d = -40\,, \ ## into your equation, I get ##\ v_1=5.7 \rlap{\;/////}-5.7\, . \ ## (The units being the same as you specified.)
EDIT: Sorry for the above typo. I actually got ##\ \ v_1=5.7\ \rm{m\, s^{-1}}##.​
Thanks: @A.T. and @kuruman !​

Solving your equation for ##v_1## first and then making the substitutions, I get the same result.

By the way, using ##\LaTeX## to format your equation, it is displayed as follows.

##\Delta d = v_1\cdot t + \frac{1}{2} a t^2## .

Last edited:
cherryLoop said:
I solved for the inital velocity by using delta d = v1xt + 1/2at^2, and using 3.5 seconds as time, acceleration as -9.8 and delta d as -40 meters. The answer I got was -5.7 m/s
SammyS said:
Plugging your values, a=−9.8, t=3.5, and Δd=−40, into your equation, I get v1=−5.7.
Can you both show your work?

A.T. said:
Can you both show your work?
Ditto that. I disagree that the initial velocity, with up as positive, is -5.7 m/s.

Even better. Substitute all the numbers and see if the equation is verified.
##\Delta d=v_0t-\frac{1}{2}gt^2##
With Δd = -40 m, ##v_0 = ## -5.7 m/s, g = 9.8 m/s2, t = 3.5 s,
##-40 \overset{?}{=}(-5.7)*(3.5)-0.5*9.8*3.5^2##
##-40 \overset{?}{=}-19.95-60.025##
##-40 \neq -79.975##
The numbers don't work.

Last edited:
SammyS
kuruman said:
Ditto that. I disagree that the initial velocity, with up as positive, is -5.7 m/s.
Doh !

Right! Just a little (but important) typo. I'll make a note in editing that post.

I did get ##v_1=5.7\ \rm{m\, s^{-1}} ##

kuruman
cherryLoop said:
TL;DR Summary: I need help with solving a kinematics problem, thanks!

Brynn is standing on a balcony that is 40 m above the ground. She is surprised by a pigeon and throws a flower pot up in the air. It takes 3.5 s for the flowerpot to smash to the ground below. The flower pot experiences acceleration due to gravity of 9.8 m/s2[down]. Determine the initial velocity of the flower pot when Brynn released it and how fast the flowerpot was moving just before it smashed to the ground.

This question has been confusing me for a long time. I solved for the inital velocity by using delta d = v1xt + 1/2at^2, and using 3.5 seconds as time, acceleration as -9.8 and delta d as -40 meters. The answer I got was -5.7 m/s [up] but it confuses me because can the acceleration upwards be less than the acceleration downwards? And I'm not sure if my answer is correct. Could someone help correct me? Thank you so much.
Edit: Thanks to everyone's response. To those asking for my work, I attached a file down below. I said that the second velocity would be up because I set the position as up, meaning that acceleration of gravity, which has the direction of down, is now a negative value, because I set the general direction as up to stay consistent. I'm not sure I see the error that everyone is talking about. Please let me know. Thanks everyone!

It's a matter of careless algebra. The problem is here

The top equation is fine. The bottom equation does not follow from it. Can you spot what went wrong?

jbriggs444
Transcribing the pictured working...

##\vec{\Delta d} = -40 \text{ m}## Yes. The displacement is 40 meters downward or -40 meters upward.
##\vec{a} = -9.8 m/s^2## Yes. The acceleration is 9.8 m/s2 downward or -9.8 m/s2 upward.
##\Delta t = 3.5 \text{ sec}## Yes.
You solve for the initial velocity ##v_1## using the SUVAT equation: ##\vec{\Delta d} = \vec{v_1} (\Delta t) + \frac{1}{2}\vec{a}(\Delta t)^2##

You plug in the numbers (prematurely). You should do algebra first and plug in numbers last. But let us proceed...

And @kuruman has spotted the problem before I got to it. So it goes.

Last edited:
kuruman
jbriggs444 said:
So it goes.
To paraphrase Leroy "Satchel" Paige, you win some, you lose some, some get locked, but you gotta reply to all of them.

jbriggs444
Zam said:
### Summary
- **Initial Velocity (v0)**: -1.99 m/s
- **Final Velocity (v)**: -1.99 m/s
How can the initial and final velocities be the same, since flower pot started off at the balcony and ended up on the ground? Besides that, how can both velocities be negative?

I haven't worked the problem, but others have done so and arrived at significantly different answers.

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