Kinematics:Interpreting graphs and deriving equations(Check my solutions please)

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SUMMARY

The discussion focuses on a kinematics problem involving a flowerpot thrown by Marian at an initial velocity of 2.1 m/s, which takes 3 seconds to hit the ground under the influence of gravity (9.8 m/s²). The calculations confirm that the height of Marian's balcony is approximately -37.82 meters, indicating a misunderstanding in the sign convention, as height cannot be negative. The final velocity of the flowerpot just before impact is calculated to be -27.3 m/s. The third equation of motion, Δd=V1Δt - 1/2 aΔt², is clarified as being suitable for upward motion scenarios, particularly when dealing with negative acceleration.

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Homework Statement


Marian who is standing on her balcony is surprised by a pigeon, and throws a flowerpot up, in the air at 2.1 m/s. It takes a total of 3 s for the flowerpot to smash to the ground. The flowerpot experiences acceleration due to gravity of 9.8m/s

a) How high is Marians balcony?
b)How fast was the flowerpot moving just before it smashed into the ground?

Homework Equations



Δd=V1Δt + 1/2 aΔt^2

V2=V1+aΔt

Δd=V1Δt - 1/2 aΔt^2
(By the way id like to know what this third equation is for, when is it appropriate to use it? I know when to use it when i see a v-t graph and am required to find displacement, and geometrically there are 2 objects in graph, i have to find area, of one and to do so would require me to subtract the area from other geometrical object, in such a scenario i would use the 3rd equation, but if the quest is in written word problem format, i don't know when to use the 3rd formula.)


The Attempt at a Solution



a) v=2.1m/s

Δt=3 s

a=-9.8m/s

Δd=V1Δt + 1/2 aΔt^2

=(2.1)(3)+1/2(-9.81)(3)^2

=6.3+1/2(-9.81)(9)

=6.3+1/2-88.79

=6.3-44.14

Δd=-37.82m



b) v1=2.1m/s

Δt=3s

a=-9.8m/s


v2=v1+aΔt

=2.1+(-9.8)(3)

V2=-27.3 m/s



Do these answers/solutions make sense?
 
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hi supernova1203! :wink:

yes that all looks fine :smile:
supernova1203 said:
Δd=V1Δt + 1/2 aΔt^2

V2=V1+aΔt

Δd=V1Δt - 1/2 aΔt^2
(By the way id like to know what this third equation is for, when is it appropriate to use it? …

my recommendation is that you should always use the first one, if necessary of course using a negative number for a
 
as far as i know the 3rd equation is used for upward motions to make it a bit easy cause others usually forget to use a negative sign for the magnitude or amount for "a" when using the 1st equation that you stated earlier :]

so i think its better to use the 3rd one when you know it the object is thrown upwards or something. to just substitute it directly though it may be confusing at times but you'll get used to it.

and yes they are all fine :)
 

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