Homework Help: SUVAT motion question - Ball thrown vertically

1. Sep 28, 2011

TheGreenMarin

1. The problem statement, all variables and given/known data
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

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

3. The attempt at a solution

s=?
u=?
v=?
a=9.81ms^-2 (negative travelling 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.

2. Sep 28, 2011

Uniquebum

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.

3. Sep 28, 2011

Staff: Mentor

There are two equations you can use.
First fill in these:
s = ?
v = ?
(You have enough information to fill in these values.)

4. Sep 28, 2011

grzz

THINK!!!
What is the velocity at the max height?

5. Sep 28, 2011

TheGreenMarin

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?]

6. Sep 28, 2011

TheGreenMarin

0 velocity?

7. Sep 28, 2011

TheGreenMarin

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

8. Sep 28, 2011

Staff: Mentor

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.)

9. Sep 28, 2011

TheGreenMarin

Ah, thankyou s now makes sense

10. Sep 28, 2011

TheGreenMarin

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. Sep 28, 2011

Staff: Mentor

OK, except that you have the wrong sign for the acceleration. Fix that first.

No it doesn't. Try again.

12. Sep 28, 2011

TheGreenMarin

-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. Sep 28, 2011

Staff: Mentor

Good.
Show how you got that, step by step.

14. Sep 28, 2011

TheGreenMarin

0=u x 4.2 + 0.5 x -9.81 x 4.2^2

so 0 = u x -82.3242

15. Sep 28, 2011

Staff: Mentor

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...

16. Sep 28, 2011

TheGreenMarin

Just confused on where the - sign has come from in the line: 0 = 4.2u -(0.5)(9.81)(4.2^2)

17. Sep 28, 2011

Staff: Mentor

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. Sep 28, 2011

TheGreenMarin

4.2u = 86.5242

/4.2 = 20.601 ms^-1 Final answer for (a) I think?

19. Sep 28, 2011

TheGreenMarin

Right OK, I see how you have done that. Is my answer of 20.601 correct?

20. Sep 28, 2011

Staff: Mentor

Yes. (I'd round it off to no more than 3 sig figures.)