Equation of uniformly accelerated motion

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Homework Help Overview

The discussion revolves around a problem involving the equation of motion for a ball thrown vertically upwards with an initial velocity. Participants are exploring the effects of acceleration due to gravity on the ball's velocity and displacement over time.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the equation v = u + at to determine the final velocity after 2 seconds, with some expressing confusion about the signs of the variables involved.
  • There is a debate regarding the displacement of the ball, with differing opinions on whether it is zero or a positive value based on the ball's motion.
  • Some participants question the units of acceleration and clarify the relationship between different notations.

Discussion Status

The discussion is active, with participants providing insights and corrections regarding the signs in the equations and the interpretation of displacement. There is no clear consensus on the displacement value, as multiple interpretations are being explored.

Contextual Notes

Participants are navigating through the implications of vertical motion and the effects of gravity, with some acknowledging potential errors in their reasoning or calculations. There is a mention of a discrepancy between personal calculations and an answer book's value for displacement.

fcb
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Homework Statement



A ball is thrown vertically upwards at 16ms^-1. What is the velocity after 2 seconds
(acceleration is 9.8 m/s/s)

Homework Equations


v = u+at
V=velocity
u= Initial speed
a=acceleration down to earth
t=Time

The Attempt at a Solution



I don't quite understand it.
 
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v = u + at

v = final velocity, u = initial velocity, a = acceleration (i.e. gravity of 9.8 m/s^-1 downwards), t = time

v = 16 + (-9.8)(2)
= -3.6 m/s^-1, i.e. the ball is now traveling downwards.
 
dink87522 said:
v = u + at

v = final velocity, u = initial velocity, a = acceleration (i.e. gravity of 9.8 m/s^-1 downwards), t = time

v = 16 + (-9.8)(2)
= -3.6 m/s^-1, i.e. the ball is now traveling downwards.

Thanks so much

You're a life saver.
 
Ohh and is the displacement 51.6m?
 
I could be wrong on this, but wouldn't the displacement be 0? There is no motion in the horizontal plane, there is only motion in the vertical plane and the ball returns to its inital starting position, hence displacement is 0.
 
dink87522 said:
I could be wrong on this, but wouldn't the displacement be 0? There is no motion in the horizontal plane, there is only motion in the vertical plane and the ball returns to its inital starting position, hence displacement is 0.

Well the answer book said 12.4m

Using the equation: r=ut + 1/2at^2
r = displacement
u = Initial
t = time
a = acceleration
 
fcb said:
Ohh and is the displacement 51.6m?

Hi fcb! :smile:

You've got a sign wrong … this u and a have opposite signs. :wink:
 
tiny-tim said:
Hi fcb! :smile:

You've got a sign wrong … this u and a have opposite signs. :wink:

I thought to myself that can't be right because gravity is 9.8m/s/s relative. So maximum number in 2 seconds can only be 19.6m/s

thanks heaps lad.
 
@tiny-tim : Well his equation is kind of correct, as in fact it's when you substitute acceleration that the - sign comes up. Though you can easily forget to substitute the sign too.

@dink87522: You kinda confused acceleration units there, it's ms-2 instead of ms-1. I know it's a mistake though, don't worry.
 
  • #10
Is m/s/s the same as ms^-2?
 
  • #11
fcb said:
Is m/s/s the same as ms^-2?

(try using the X2 tag just above the Reply box :wink:)

Yes. :smile:
 
  • #12
I am on my phone so I can't use it. Well not I currently know of
 
  • #13
ah! you can type [noparse]"" and "" instead.[/noparse] :wink:
 

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