Kinimatics: Find a & b for Projected Particle

[Taylor_1989]

I have gotten the right ans for this question, but I am not too sure if I am worked it out the right way.

Question: A particle is projected vertically upwards from a point O with speed $u ms^{-1}$. Two seconds later it is still moving upwards with a speed $\frac{1}{3}u ms^{-1}$. Find a: the value of u, b: the time from instant that the particle leaves O to the instant that it returns to O.

I worked out U which is 29.4ms^-1.

It b: I am having trouble with, as I stated I have the right ans, but to sure on if I have gotten the ans correctly.

What I have done it is take the part where the ball is at 1/3 u and workout out the time of travel from the part where it passes 1/3u again and the add on 4 seconds. Not very good at explain hopefully the math will show better.

$0=9.8t-4.9t^2 → 4.9t^2-9.8t=0$ solve and t=0 and t=2. So I then add on the 4sec and total time is 6 secs. Is this correct or not?

I would appreciate any help, big thanks in advance.

Sorry for the repost, the first one keeps telling me its invalid I can't edit it at all. So could one of the mods remove my older post.

 

[Doc Al]

$0=9.8t-4.9t^2 → 4.9t^2-9.8t=0$ solve and t=0 and t=2. So I then add on the 4sec and total time is 6 secs. Is this correct or not?

Please explain where this equation comes from. Where did you get a velocity of 9.8?

Why not deal directly with velocity equations? Much simpler.

 

[Taylor_1989]

I got it from the 1/3*29.4. I didn't know that I could deal with directly as there are two different velocity, for the way up. That is why I broke it down.

 

[Doc Al]

I got it from the 1/3*29.4.

But that kind of assumes the answer, doesn't it?

I didn't know that I could deal with directly as there are two different velocity, for the way up. That is why I broke it down.

Just use v_f = v_i - gt .

 

[Taylor_1989]

Okay I am with you. Thanks for the help.