## calculusxy one year ago MEDAL!!! A ball is thrown straight up. What will be the instantaneous velocity at the top of its path? What will be its acceleration at the top? Why are your answers different?

1. calculusxy

@jim_thompson5910

2. anonymous

Is it not that when it goes upward, at the top of its path, velocity =0?

3. jim_thompson5910

OOOPS is correct Gravity is pulling the ball down. The initial throw pushes the ball up and the velocity decreases as the ball climbs higher because it's fighting gravity. Once the ball gets to the peak, the velocity hits 0. The acceleration due to gravity is always the same no matter how fast the ball is moving.

4. calculusxy

What would be its acceleration?

5. jim_thompson5910

The acceleration due to gravity is always the same no matter how fast the ball is moving. acceleration of gravity = 9.8 m/s^2 = 32 ft/s^2 (both are approximate)

6. calculusxy

How did you calculate that 9.8m/s^2 = 38ft?s^2 ?

7. jim_thompson5910

the acceleration of gravity only changes when the height is drastically different technically there is change, but it's so small that it's hard to notice

8. jim_thompson5910

it's just something you look up in a table, book, or something. or you memorize it

9. calculusxy

I mean i know that 9.8m/s^2 is the instantaneous speed of a free-falling object. but how did you just calculate the acceleration at the top? acceleration = gravity x time ?

10. jim_thompson5910

the acceleration is equal to the acceleration of gravity

11. calculusxy

Honestly, i have no idea to what you just said.

12. jim_thompson5910

imagine the object in free fall every second, the speed increases by 9.8 m/s

13. calculusxy

okay

14. jim_thompson5910

let t be the time elapsed since dropping the object and letting it free fall (ignore air resistance) t = 0 speed = 0 m/s t = 1 speed = 9.8 m/s t = 2 speed = 19.6 m/s (9.8 times 2 = 19.6) t = 3 speed = 29.4 m/s (9.8 times 3 = 29.4) etc etc

15. calculusxy

yeah...

16. calculusxy

so what's going to happen next?

17. jim_thompson5910

No matter where the object is, it's feeling the same force on it. So it undergoes the same acceleration throughout its journey going up and then coming back down

18. zzr0ck3r

acceleration is a vector. i.e. it has direction.

19. calculusxy

but my question was how did you get this: "acceleration of gravity = 9.8 m/s^2 = 32 ft/s^2 (both are approximate)"

20. zzr0ck3r

Isac Newton

21. zzr0ck3r

I think...

22. jim_thompson5910

Galileo discovered it long before Newton, but yeah pretty much

23. zzr0ck3r

did he discover the speed or just the thing?

24. jim_thompson5910

those two numbers are constants that you memorize or look up

25. calculusxy

so the acceleration is 32?

26. jim_thompson5910

32 ft per s^2, yes

27. jim_thompson5910

just the speed/acceleration I think @zzr0ck3r I don't think he actually came up with unifying theories like Newton did

28. zzr0ck3r

And almost got killed as a result. Ahh humans...

29. calculusxy

so i found this website where it said that the acceleration can just be defined as "g", i believe:

30. zzr0ck3r

yes. gravity is constant and we normally just say g because we are estimating it when we give an actually number.

31. zzr0ck3r

like $$\pi$$ and 3.14. They are not the same but we use 3.14 when we need a close number.

32. jim_thompson5910

g = 9.8 m/s^2 or g = 32 ft/s^2 both are the same. Saying 'g' is a shorthand way to say "acceleration of gravity on earth at sea level"

33. jim_thompson5910

When you hear things like "the rocket is going 3 g's (idk I made up the number)" it means "the rocket is accelerating 3 times the acceleration of gravity" so the rocket's acceleration would be 29 m/s^2

34. calculusxy

Thank you. I have another question.

35. phi

If you have time and are interested, see https://www.khanacademy.org/science/physics/newton-gravitation/gravity-newtonian/v/introduction-to-gravity

36. calculusxy

If a salmon swims straight upward in the water fast enough to break through the surface at a speed of 5m/s, how high can it jump above the water?

37. jim_thompson5910

Acceleration of gravity: g = 9.8 m/s^2 Initial Velocity Vo = 5 m/s Plug those values into h = -(g/2)t^2 + Vo*t We don't know anything about the angle in which it jumps, so ignore that. Assume it jumps straight up. Find the vertex to get the answer.

38. calculusxy

h = (-9.8/2)5^2 + 5*2

39. jim_thompson5910

t is unknown for now

40. calculusxy

Yeah that's what I was guessing

41. jim_thompson5910

you should get h = -4.9t^2 + 5t find the vertex of this to figure out the highest point

42. calculusxy

How would I figured out the vertex?

43. calculusxy

*figure

44. jim_thompson5910

h = -4.9t^2 + 5t is in the form h = at^2 + bt + c a = -4.9 b = 5 c = 0 plug the values of 'a' and 'b' into -b/(2a) to figure out the time value at the peak once you get this value, plug it into h = -4.9t^2 + 5t to find h

45. calculusxy

so t would be -b/(2a) ?

46. jim_thompson5910

the t value at the peak, yes |dw:1436315387333:dw|

47. calculusxy

Okay

48. calculusxy

-5/(2 x (-4.9))

49. jim_thompson5910

yes

50. calculusxy

-5/(-9.8) = 0.51

51. jim_thompson5910

then you plug t = 0.51 into h = -4.9t^2 + 5t

52. calculusxy

-4.9(0.51)^2 + 5(0.51) -4.9(0.2601) + 5(0.51) 1.27449 + 2.55 3.82449

53. jim_thompson5910

54. calculusxy

Oh yea

55. calculusxy

1.27551

56. jim_thompson5910

Yep, 1.27551 the max height is roughly 1.27551 meters and this happens at around 0.51 seconds

57. calculusxy

Thank you! This is a worksheet, so can u make sure all of my answers? I will post them in another post.

58. calculusxy

It's like 5 questions?

59. jim_thompson5910

ok make a brand new post