So, V_F= 11m/sWhat is the speed of the potted plant after falling 6m?

  • Thread starter Thread starter mcHefner
  • Start date Start date
  • Tags Tags
    Plant Speed
AI Thread Summary
The discussion centers on calculating the final speed of a potted plant after falling 6 meters from a 12-meter high ledge. Participants agree that mass is irrelevant in this scenario due to the conservation of mechanical energy. Using the equation v² = 2gh, the first participant calculates a final speed of approximately 11 m/s. Another participant arrives at a slightly lower speed of about 10.84 m/s using a different method involving initial velocity and acceleration. Both calculations confirm the principles of energy conservation in free fall.
mcHefner
Messages
1
Reaction score
0

Homework Statement


[PLAIN]http://img8.imageshack.us/img8/664/photo4dc.jpg
Sorry it is sideways.

Only known data is the 6m drop from the 12m high ledge.

Homework Equations



Currently wokring with Kinetic energy (1/2mv^2), potential energy(mgh).

The Attempt at a Solution



My thinking is mass doesn't matter, it will cancel out due to the mechanical energy( if that makes sense). Therefore the equation to solve is v2f=2gh

Therefore after subbing in 9.81m/ss for g, 6m for h; my final answer is 11m/s with rounding because of sig figs.

Thank you for help
 
Last edited by a moderator:
Physics news on Phys.org
Uh, do you think you could make that figure a little bigger ... I mean, I can almost see 25% of it at a time as it is.
 
I also got 11m/s, but found that answer in a less complicated way:

You know:
velocity_initial is 0m/s
Distance is 6m
Acceleration is 9.8m/ss

So:
velocity_final= SQRT(2*A*D)
V_F= SQRT(2*9.8*6)
V_F= 10.844m/s
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

What is the Recoil Speed and Energy Stored in a Spring Gun After a Collision?
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top