## mayankdevnani 3 years ago A ball is dropped on to the floor from a height of 10 m. It reboundes to a height of 2.5 m. If the ball is in contact with the floor for 0.01 seconds, what is the average acceleration during contact? a) \[1400 m/s^2\] b) \[2100 m/s^2\] c) \[700 m/s^2\] d) \[2800 m/s^2\]

1. mayankdevnani

@amistre64 @hba @asnaseer @AccessDenied @jhonyy9 @jazy @ganeshie8

2. tictac4456

The acceleration of a body is defined as the change of its velocity during an interval of time divided by the interval of time. Expressed algebraically, a = (Vo - Vf)/t where a = the acceleration, Vo = the initial velocity, Vf = the final velocity and t = the interval of time.

3. tictac4456

From the basic a = (Vo - Vf) we derive the first as ......................................t ..............................Vf = Vo + at (the acceleration is assumed constant)

4. tictac4456

The second defines the relationship between the distance traveled by a body exposed to a constant acceleration. The distance traveled over an interval of time derives from the product of the average velocity during the time interval and the time interval. The average velocity is (Vo + Vf)/2 and the time interval "t" which yields d = (Vo + Vf)t/2. Substituting the Vf derived above, d = [Vo + (Vo + at)]t/2 resulting in ...............................d = Vo(t) + a(t^2) (quite often, s is used in place of d) ......................................................2

5. tictac4456

The third combines the first two by eliminating the time interval "t". Therefore, a(d) = (Vo - Vf)(Vo + Vf)t = (1/2)(Vf - Vo)(Vf + Vo) resulting in .....................................t.............2 ..............................Vf^2 = Vo^2 + 2ad

6. mayankdevnani

thnx.... @tictac4456

7. tictac4456

no problem do u go to connexus

8. mayankdevnani

is it right?? While falling down, initial velocity = 0, final velocity = v, acceleration due to gravity = 9.8 m/s 2 , height = 10 m The velocity of the ball with which it hits the ground can be found using, v 2 = u 2 + 2as => v 2 = 0 + 2×9.8×10 => v = 14 m/s (downward) While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = -9.8 m/s2 [this time the velocity and acceleration are oppositely directed so we have the negative sign] , height = 2.5 m The velocity of the ball with which it leaves the ground can be found using, v 2 = u 2 + 2as => 0 = u 2 - 2×9.8×2.5 => u = 7 m/s (upward) So, Change in velocity = 7 m/s (upward) – 14 m/s (downward) => Δv = 7 m/s + 14 m/s [both in upward direction] => Δv = 21 m/s So, acceleration is, a = Δv/Δt = 21/0.01 = 2100 m/s 2

9. tictac4456

yup

10. mayankdevnani

@tictac4456

11. tictac4456

thats correct

12. mayankdevnani

thnx..

13. mayankdevnani

@satellite73 help me:)

14. mayankdevnani

@amistre64

15. tictac4456

hey ur work is correct but the answer is wrong

16. tictac4456

17. mayankdevnani

howz that??

18. tictac4456

oh im srry never mind i didnt see the seconds ur answer is correct

19. tictac4456

im 100% sure its right

20. mayankdevnani

ohkkk!!

21. asnaseer

22. mayankdevnani

thnx...

23. asnaseer

yw :)

24. tictac4456

do ya take entrepuener class

25. mayankdevnani

what do you mean??

26. tictac4456

other than physics do u take entrepuener class

27. tictac4456

28. mayankdevnani

no!! never!!! why??

29. tictac4456

haha i kinda needed help

30. tictac4456

but never mind

31. mayankdevnani

plz you can!!!

32. tictac4456

ok lol but if u dont knw it u dont have to help k

33. mayankdevnani

k

34. tictac4456

what is gross profit? A)the money left over after the cost of making a product or providing a service B) the cost of labor to produce a product C) the difference between a product's price and the revenue of the company D) a negative net profit, in which company experiences a loss

35. mayankdevnani

(finance) the net sales minus the cost of goods and services sold.

36. tictac4456

thnkz lol

37. tictac4456

i gave u a medal right

38. mayankdevnani

right!!!

39. tictac4456

cool so did u graduate? or u still in high skool

40. mayankdevnani

high school!!!

41. tictac4456

42. mayankdevnani

9 class

43. mayankdevnani

ok @tictac4456

44. tictac4456

me2 lol

45. tictac4456

what skool?