## anonymous one year ago A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes time 0.420 s to pass this window, which is of height 1.90 m. Question- How far is the top of the window below the windowsill from which the flowerpot fell?

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1. anonymous

@UnkleRhaukus

2. anonymous

@johnweldon1993

3. UnkleRhaukus

|dw:1442325952193:dw|

4. UnkleRhaukus

what are the forces acting on the flower pot during the motion?

5. anonymous

you mean the gravity?

6. anonymous

Take the free fall acceleration to be 9.80m/s^2 .

7. UnkleRhaukus

yeah gravity is the only force acting here!

8. UnkleRhaukus

|dw:1442326446681:dw|

9. UnkleRhaukus

by newtons2nd the sum of the forces is equal the mass times its acceleration ∑ F = ma = mg (if we define the y axis pointing downwards ) so a = g

10. UnkleRhaukus

we get y(t) = y_0 + v_y0 t + 1/2 g t^2

11. anonymous

hmmm..can you show me how you derived to that equation

12. anonymous

the one i know is d=Vit + 1/2at^2

13. UnkleRhaukus

a = y'' = g v = y' = ∫ g dt = gt +c y = ∫ (gt + c) dt = 1/2 gt^2 +c_1t + c_2

14. UnkleRhaukus

y(0) = y_0 --> c_2 = y_0 v(0) = v_y0 --> c_1 = v_y0

15. anonymous

ok...how can we solve this ^^ using that formula

16. UnkleRhaukus

y(t_1) = y_0 + v_y0 (t_1) + 1/2 g (t_1)^2 y(t_2) = y_0 + v_y0 (t_2) + 1/2 g (t_2)^2 D = y(t_2) - y(t_1) = v_y0 (t_2-t_1) + 1/2 g (t_2-t_1)^2 is this like your formula?

17. anonymous

yup

18. UnkleRhaukus

so D is the displacement? and (t_2 - t_1) is the time it takes of the pot to fall past the window

19. anonymous

absolutely

20. UnkleRhaukus

how much displacement is is this example?

21. anonymous

1.90m

22. UnkleRhaukus

so we have ( v_i = v_y0 ) D = v_i t + 1/2 g t^2 lets solve this equation for v_i, the velocity of the pot plant when it was falling past the top the the window and Then, calculate with D = 1.90 m t = 0.420 s g = 9.80 m/s^2

23. UnkleRhaukus

what do you get when you rearrange the equation ?

24. anonymous

2.4658

25. anonymous

m/s

26. anonymous

is that we only need?

27. UnkleRhaukus

now we have the initial velocity , we still need to work out the height to the window sill

28. UnkleRhaukus

Do you have formulas you can just use, or is the derivation helpful?

29. anonymous

hmmm.. idk which formula to use lol

30. UnkleRhaukus

we need to find the distance to the windowsill from the lower window, we have the velocities

31. UnkleRhaukus

do you have a set of 5 equations of projectile motion?

32. UnkleRhaukus

or i suppose we can say the the pot started on the sill at rest, and was falling at 2.4658 m/s, at the top on the next window down how long was it falling ?

33. anonymous

we dont learn projectile from now so i really dont know

34. anonymous

@rishavraj