Equation of uniformly accelerated motion

AI Thread Summary
A ball is thrown vertically upward at 16 m/s, and after 2 seconds, its velocity is calculated to be -3.6 m/s, indicating it is moving downward due to gravity's acceleration of 9.8 m/s². There is confusion regarding the displacement, with one participant suggesting it is 0 since the ball returns to its starting position, while the answer book states it is 12.4 m. The discussion highlights the importance of correctly applying signs in the equations of motion, particularly when considering acceleration and initial velocity. Participants clarify that acceleration should be expressed as m/s², not m/s. Overall, the thread emphasizes the correct application of kinematic equations in uniformly accelerated motion.
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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