calculusxy
  • calculusxy
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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calculusxy
  • calculusxy
@jim_thompson5910
anonymous
  • anonymous
Is it not that when it goes upward, at the top of its path, velocity =0?
jim_thompson5910
  • 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.

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calculusxy
  • calculusxy
What would be its acceleration?
jim_thompson5910
  • 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)
calculusxy
  • calculusxy
How did you calculate that 9.8m/s^2 = 38ft?s^2 ?
jim_thompson5910
  • 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
jim_thompson5910
  • jim_thompson5910
it's just something you look up in a table, book, or something. or you memorize it
calculusxy
  • 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 ?
jim_thompson5910
  • jim_thompson5910
the acceleration is equal to the acceleration of gravity
calculusxy
  • calculusxy
Honestly, i have no idea to what you just said.
jim_thompson5910
  • jim_thompson5910
imagine the object in free fall every second, the speed increases by 9.8 m/s
calculusxy
  • calculusxy
okay
jim_thompson5910
  • 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
calculusxy
  • calculusxy
yeah...
calculusxy
  • calculusxy
so what's going to happen next?
jim_thompson5910
  • 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
zzr0ck3r
  • zzr0ck3r
acceleration is a vector. i.e. it has direction.
calculusxy
  • 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)"
zzr0ck3r
  • zzr0ck3r
Isac Newton
zzr0ck3r
  • zzr0ck3r
I think...
jim_thompson5910
  • jim_thompson5910
Galileo discovered it long before Newton, but yeah pretty much
zzr0ck3r
  • zzr0ck3r
did he discover the speed or just the thing?
jim_thompson5910
  • jim_thompson5910
those two numbers are constants that you memorize or look up
calculusxy
  • calculusxy
so the acceleration is 32?
jim_thompson5910
  • jim_thompson5910
32 ft per s^2, yes
jim_thompson5910
  • jim_thompson5910
just the speed/acceleration I think @zzr0ck3r I don't think he actually came up with unifying theories like Newton did
zzr0ck3r
  • zzr0ck3r
And almost got killed as a result. Ahh humans...
calculusxy
  • calculusxy
so i found this website where it said that the acceleration can just be defined as "g", i believe:
zzr0ck3r
  • zzr0ck3r
yes. gravity is constant and we normally just say g because we are estimating it when we give an actually number.
zzr0ck3r
  • zzr0ck3r
like \(\pi\) and 3.14. They are not the same but we use 3.14 when we need a close number.
jim_thompson5910
  • 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"
jim_thompson5910
  • 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
calculusxy
  • calculusxy
Thank you. I have another question.
phi
  • phi
If you have time and are interested, see https://www.khanacademy.org/science/physics/newton-gravitation/gravity-newtonian/v/introduction-to-gravity
calculusxy
  • 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?
jim_thompson5910
  • 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.
calculusxy
  • calculusxy
h = (-9.8/2)5^2 + 5*2
jim_thompson5910
  • jim_thompson5910
t is unknown for now
calculusxy
  • calculusxy
Yeah that's what I was guessing
jim_thompson5910
  • jim_thompson5910
you should get h = -4.9t^2 + 5t find the vertex of this to figure out the highest point
calculusxy
  • calculusxy
How would I figured out the vertex?
calculusxy
  • calculusxy
*figure
jim_thompson5910
  • 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
calculusxy
  • calculusxy
so t would be -b/(2a) ?
jim_thompson5910
  • jim_thompson5910
the t value at the peak, yes |dw:1436315387333:dw|
calculusxy
  • calculusxy
Okay
calculusxy
  • calculusxy
-5/(2 x (-4.9))
jim_thompson5910
  • jim_thompson5910
yes
calculusxy
  • calculusxy
-5/(-9.8) = 0.51
jim_thompson5910
  • jim_thompson5910
then you plug t = 0.51 into h = -4.9t^2 + 5t
calculusxy
  • calculusxy
-4.9(0.51)^2 + 5(0.51) -4.9(0.2601) + 5(0.51) 1.27449 + 2.55 3.82449
jim_thompson5910
  • jim_thompson5910
1 Attachment
calculusxy
  • calculusxy
Oh yea
calculusxy
  • calculusxy
1.27551
jim_thompson5910
  • jim_thompson5910
Yep, 1.27551 the max height is roughly 1.27551 meters and this happens at around 0.51 seconds
calculusxy
  • calculusxy
Thank you! This is a worksheet, so can u make sure all of my answers? I will post them in another post.
calculusxy
  • calculusxy
It's like 5 questions?
jim_thompson5910
  • jim_thompson5910
ok make a brand new post

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