Solving for Initial Velocity & Max Height of Thrown Ball

[quicksilver123]

Homework Statement

A ball is thrown upwards. Its velocity at 3/5 its max height is 16.0m/s
What is its initial velocity?
What is its max height?

Homework Equations

vf^2 = vi^2 -2g(yf-yi)

define up as positive

The Attempt at a Solution

I ran through numerous attempts on paper but can't seem to find a proper answer. The closest I got was an answer that spit out a negative height... help? Somewhat urgent as I need to bring this question in tomorrow for homework.

Thank you![/B]

 

[gneill]

Show your attempt, wrong or right.

 

[quicksilver123]

Sigh. I crossed it all out on my paper, otherwise I'd just take a picture and scan it.

0=(16m/s)^2 -2(-9.8m/s/s)(2h/5)
-(16m/s)^2=18.6m/s/s(2h/5)
-(16m/s)^2= (7.44h)m/s/s
-256m^2/s^2 = h(7.44)m/s/s

Two issues here being that
a) the units do not match since I haven't multiplied in height, so operations may be difficult
b) there is a negative, so the height will come out as a negative, which seems illogical as I have specified up as positive

 

[gneill]

Sigh. I crossed it all out on my paper, otherwise I'd just take a picture and scan it.

0=(16m/s)^2 -2(9.8m/s/s)(2h/5)
-(16m/s)^2=-18.6m/s/s(2h/5)

Why didn't you cancel the minus signs on both sides here?

-(16m/s)^2= (7.44h)m/s/s
-256m^2/s^2 = h(7.44)m/s/s

Two issues here being that
a) the units do not match since I haven't multiplied in height, so operations may be difficult

Looks to me like m/s² on the right will cancel out the same on the left leaving m. Thus the units of h will be m, which is correct.

b) there is a negative, so the height will come out as a negative, which seems illogical as I have specified up as positive

Nope. You just didn't cancel the signs at a previous step.

 

[quicksilver123]

No, that was a typo. I fixed it. Stated in the original post that up was to be positive. Therefore the acceleration is negative.
Thank you for your comment on the units but the height still appears to turn out negative.

 

[gneill]

No, that was a typo. I fixed it. Stated in the original post that up was to be positive. Therefore the acceleration is negative.
Thank you for your comment on the units but the height still appears to turn out negative.

Then you've introduced an excess negative by both subtracting the term 2gh and making the constant g negative.

You can either treat g as a positive constant and account for the direction in the algebra, or you can make "2ad" a positive term and let g be negative to account for the direction.

The general form of the equation is:

$v_f^2 - v_i^2 = 2 a d$

so then:

$-v_i^2 = -v_f^2 + 2 a d$
$v_i^2 = v_f^2 - 2 a d$

and if $v_f = 0$ and $a = g$ is a negative value,

$v_i^2 = -2 g d$

will yield a positive result for d.

 

[quicksilver123]

Thank you!
I have a result for height. However, attempting to find the initial velocity... my result is the same as the given velocity (when the ball is a fraction of the max height).

I used the same equation and substituted height.

 

[gneill]

Thank you!
I have a result for height. However, attempting to find the initial velocity... my result is the same as the given velocity (when the ball is a fraction of the max height).

I used the same equation and substituted height.

Can you show your work?

 

[quicksilver123]

Sorry, it was my mistake. I solved the question. thank you very much for your help.

 

[gneill]

Okay! Cheers.