SUVAT motion question - Ball thrown vertically

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Homework Help Overview

The discussion revolves around a physics problem involving the motion of a cricket ball thrown vertically upwards, with a total time of 4.2 seconds for the ball to return. Participants are tasked with finding the initial velocity and maximum height using SUVAT equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of SUVAT equations, questioning how to assign values to the variables involved. Some suggest starting with the equation v = v_0 + at, considering the velocity at the maximum height to be zero. Others express uncertainty about the displacement and the signs of acceleration.

Discussion Status

The discussion has progressed with participants attempting to apply various equations and sharing their calculations. Some have provided guidance on correcting signs and rearranging equations, while others are still grappling with the implications of their findings and the correct application of the equations.

Contextual Notes

Participants note the challenge of determining the correct values for variables and the implications of negative results in their calculations. There is an acknowledgment of the need to clarify assumptions regarding the motion and the effects of gravity.

TheGreenMarin
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Homework Statement


A cricket ball is thrown vertically upwards and returns in 4.2s. Find (a)The initial velocity of the ball

(b)The maximum height reached


Homework Equations


v=u+at
v^2=u^2+2as
s=(u+v)/2 multiplied by t
s=ut+1/2at^2


The Attempt at a Solution



s=?
u=?
v=?
a=9.81ms^-2 (negative traveling upwards)
t=4.2s

I could not start the question because to use SUVAT equations at least 3 of the 5 letters must have values to be put into a equation.
 
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Start with

v = v_0 + at
Where v = end velocity, v_0 initial velocity, a = g = gravity and t = time

If you consider v to be the velocity at the top of the throw then v = 0.
Thus 0 = v_0 + at
v_0 = -at, You know the acceleration and t = half the total time. So there you get the initial velocity and rest becomes simplier.
 
There are two equations you can use.
First fill in these:
s = ?
v = ?
(You have enough information to fill in these values.)
 
THINK!
What is the velocity at the max height?
 
Doc Al said:
There are two equations you can use.
First fill in these:
s = ?
v = ?
(You have enough information to fill in these values.)

s= I'm really not sure how to found out the displacement
v=0 [ as it has stopped at its maximum point in the air?]
 
grzz said:
THINK!
What is the velocity at the max height?

0 velocity?
 
Uniquebum said:
Start with

v = v_0 + at
Where v = end velocity, v_0 initial velocity, a = g = gravity and t = time

If you consider v to be the velocity at the top of the throw then v = 0.
Thus 0 = v_0 + at
v_0 = -at, You know the acceleration and t = half the total time. So there you get the initial velocity and rest becomes simplier.

0 = v_0 + 9.81x4.2 ? I am not sure I have done this correctly

so v_0 = -41.202 <---- I've definitely gone wrong here, the initial speed surely cannot be negative
 
TheGreenMarin said:
0 = v_0 + 9.81x4.2 ? I am not sure I have done this correctly

so v_0 = -41.202
It looks like you are trying to use this equation: s=ut+1/2at^2

But you left out a few things. Try again. (But yes, s = 0 for the round trip.)
 
Doc Al said:
It looks like you are trying to use this equation: s=ut+1/2at^2

But you left out a few things. Try again. (But yes, s = 0 for the round trip.)

Ah, thankyou s now makes sense
 
  • #10
0=u x 4.2 + 0.5 x 9.81 x 4.2^2

gives me 0 = u x 90.7242

so u = 0

I'm really struggling with this question, what error have I made this time?
 
  • #11
TheGreenMarin said:
0=u x 4.2 + 0.5 x 9.81 x 4.2^2
OK, except that you have the wrong sign for the acceleration. Fix that first.

gives me 0 = u x 90.7242
No it doesn't. Try again.
 
  • #12
Doc Al said:
OK, except that you have the wrong sign for the acceleration. Fix that first.


No it doesn't. Try again.

-9.81 as its against gravity, rookie error

I now got -82.3242 as my answer but I'm not sure how this can be right as it is negative
 
  • #13
TheGreenMarin said:
-9.81 as its against gravity, rookie error
Good.
I now got -82.3242 as my answer
Show how you got that, step by step.
 
  • #14
0=u x 4.2 + 0.5 x -9.81 x 4.2^2

so 0 = u x -82.3242
 
  • #15
TheGreenMarin said:
0=u x 4.2 + 0.5 x -9.81 x 4.2^2
OK.
so 0 = u x -82.3242
Not OK.

Try this:
0=u x 4.2 + 0.5 x -9.81 x 4.2^2
which I'll rewrite as:
0 = 4.2u -(0.5)(9.81)(4.2^2)
4.2u = (0.5)(9.81)(4.2^2)

And so on...
 
  • #16
Doc Al said:
OK.

Not OK.

Try this:
0=u x 4.2 + 0.5 x -9.81 x 4.2^2
which I'll rewrite as:
0 = 4.2u -(0.5)(9.81)(4.2^2)
4.2u = (0.5)(9.81)(4.2^2)

And so on...

Just confused on where the - sign has come from in the line: 0 = 4.2u -(0.5)(9.81)(4.2^2)
 
  • #17
TheGreenMarin said:
Just confused on where the - sign has come from in the line: 0 = 4.2u -(0.5)(9.81)(4.2^2)
Starting with:
0 = 4.2u +(0.5)(-9.81)(4.2^2)

You can move that minus sign from in front of the 9.81 and place it here:
0 = 4.2u -(0.5)(9.81)(4.2^2)

(1)(2)(-3) is the same as -(1)(2)(3).
 
  • #18
4.2u = 86.5242

/4.2 = 20.601 ms^-1 Final answer for (a) I think?
 
  • #19
Doc Al said:
Starting with:
0 = 4.2u +(0.5)(-9.81)(4.2^2)

You can move that minus sign from in front of the 9.81 and place it here:
0 = 4.2u -(0.5)(9.81)(4.2^2)

(1)(2)(-3) is the same as -(1)(2)(3).

Right OK, I see how you have done that. Is my answer of 20.601 correct?
 
  • #20
TheGreenMarin said:
Right OK, I see how you have done that. Is my answer of 20.601 correct?
Yes. (I'd round it off to no more than 3 sig figures.)
 
  • #21
Doc Al said:
Yes. (I'd round it off to no more than 3 sig figures.)

Thank you very much for your help I really appreciate it. I am so glad I found this website, you have helped me greatly. I will definitely stick around here and maybe one day even help people out with their problems (Once I understand the basics).

This site is so helpful that I will tell my friends to come here if in need of any physics help.

Once again thank you Sir, you have really helped me out in understanding of SUVAT equations and of basic re-arranging.

Thank you
 
  • #22
You are most welcome!
 

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