## anonymous one year ago Physics stuff!! A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.

1. anonymous

@ganeshie8 @nincompoop @pooja195 @Preetha any ideas?

2. Empty

I believe there are two ways to do this 1) Calculate the trajectory 2) Calculate the energy change

3. ganeshie8

presuming the ball lands on the ground and not on the cliff... even though it was thrown vertically upwards

4. anonymous

this is a calc I problem by the way

5. ganeshie8

maybe the angle shouldn't matter here because the ball gains its speed by the time it reaches cliff. You may start with the given acceleration due to gravity : $v(t)-v(0)=\int\limits_0^t a(t)\,dt$

6. anonymous

okay s v(0) = 40 but how do we find v(t) ?

7. Empty

Continue further, $v(t) = \frac{d x(t)}{dt}$

8. anonymous

wait slow down , how did you find s(t) ?

9. ganeshie8

\begin{align}v(t)-v(0)&=\int\limits_0^t a(t)\,dt \\~\\ &=\int\limits_0^t -32\,dt \end{align} you're given that $$a(t)=-32$$, see if you can integrate above now..

10. anonymous

It is position. 500ft cliff above the ground=+500 Thrown upward with speed of 40 feet per second-it will eventually fall down to the ground which is negative=40t acceleration is negative because velocity is negative -32 feet per second squared t is seconds so second squared=-32t^2

11. anonymous

v(t) - 40 = (-32t^2)/2 ?

12. anonymous

Position s=0 means it is a at ground. You find t to know the time it took to reach the ground. You then used the velocity function by taking derivative of position. Plug that time it reached the ground to the velocity function.

13. ganeshie8

nope, whats antiderivative of constant ?

14. anonymous

well you add a variable in this case t

15. anonymous

oh woops i see what i did wrong

16. anonymous

v(t) = -32t + 40: also @Shalante if we take your s(t) and we derive it until we get a(t) we get -64 and i think it should be -32

17. anonymous

no it is -64

18. anonymous

$\int\limits v(t) = s(t) = -16t^2 + 40t + 500$ does this look right to you ?

19. anonymous

"se acceleration due to gravity as –32 feet per second squared"

20. anonymous

use *

21. anonymous

Actually, there lots of ways to do this. There is like 3

22. anonymous

yes that is right.

23. anonymous

i know but i have to show my work for this one so cant use physics equations sadly :(

24. anonymous

oh wait it is wrong. Supposed to be -32t^2-40t+500

25. anonymous

Make the position=0 by doing quadratic formula.

26. anonymous

use your calculator divide everything by -4 first

27. Empty

Yeah as far as physics is concerned I think this would be pretty much the best way: $K_i + V_i = K_f + V_f$ $\frac{1}{2} m 40^2 + m 32*500 = \frac{1}{2} m v^2$ $\sqrt{40^2 + 2*32*500} = v$

28. ganeshie8

careful, indefinite integral is unique only upto a constant

29. anonymous

interesting, i'm getting positive 40t because of the v(t) - v(0) part (when we added the v(0) to isolate the v(t))

30. ganeshie8

$s(t)-s(0)=\int\limits_0^t v(t) -16t^2 + 40t$ so you're right, you do get $$s(t) = -16t^2+40t+500$$

31. anonymous

okay so now just set s(t) = 0 to find when it touches the ground right?

32. anonymous

wait i thought s (t) = -16t^2 .... not -32t^2

33. ganeshie8

physicists are overjoyed by the conservation of "anything", but i think math professors still want to see the integrals mess ;)

34. anonymous

Answer is negative because the ball's speed while hitting the ground is going downwards.

35. ganeshie8

t=0 corresponds to s=500 right ?

36. ganeshie8

recall that $$s(t)$$ is the displacement from $$ground$$ after $$t$$ seconds

37. anonymous

$-32t^2+40t+500=0$ got $t=3.376s,-4.626s$ Time should only be positive so use t=3.376s $v(t)=-64t-40$ $v(3.376)=-256.06 m/s$ are you sure this is right @ganeshie8

38. anonymous

shouldn't it be $-16t^2 +40t + 500 = 0$

39. ganeshie8

sry im not following the discussion between you and Jdosio... but you're not allowed to use direct formulas here right ?

40. anonymous

no only calculus

41. anonymous

okay my only positive root for v(t) is 6.978

42. anonymous
43. anonymous

Ya use -16t^2

44. anonymous

so to find the velocity at that time we just plug that into the formula for v(t) right ?

45. anonymous

which gives me around -183.3030

46. anonymous

does that look about right to you guys?

47. anonymous

Nope it should be -40

48. anonymous

im using v(t) = -32t + 40 is that right?

49. anonymous

Answer is right, but the graph is wrong. Guess you do not need the graph.

50. anonymous

okay so then does everyone agree that the answer is -183.3030 !?

51. anonymous

Read it wrong at the beginning lol

52. anonymous

@Empty @ganeshie8 ?

53. anonymous

It is correct.

54. anonymous

okay ill stick around just incase anyone changes their mind, thanks for the help guys :D

55. Empty

I got -183.3030 by doing it with the energy considerations alone, so good job! Same answer different route is a good indicator imo.

56. ganeshie8

Yes the ball touches the ground after at t=6.978 $$v(6.978) =40-32(6.978) \approx -183.303$$ you want to avoid that negative sign though speed is $$183.303$$ whatever unit

57. Empty

Only reason why I think they'd want the negative sign there is because they gave -32 f/s^2 as the acceleration due to gravity, which imo is kind of awkward so I don't know what they want.

58. ganeshie8

acceleration is a vector, so sign makes sense speed isn't

59. anonymous

ill just make it explicit that the velocity is downwards

60. ganeshie8

61. Empty

Yeah you're right @ganeshie8 good point.

62. ganeshie8

just persisting because, when you use the terms correctly, your professor knows that you understood the concepts.. that is for @Jdosio :)

63. anonymous

ill take note of that :)

64. anonymous

okay something weird has happened

65. anonymous

i found the speed of the ball at the time that the ball is at a height of 500 (the height of the cliff) and got -40 does that look oddly round to you guys ?

66. anonymous

@ganeshie8 @Empty @Shalante ?

67. Empty

Well this is actually completely expected that the speed is the same for the same height, this is because it has no more potential or kinetic energy at this same height when you threw it. This is due to conservation of energy, which I realize is not exactly in the scope of what you're learning however it's clear at a glance to me because of this.

68. anonymous

okay so i should stick with my 183.3030 right?

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