Suvat Question - seemingly too little info

[TicTac2]

Hi all,
I am new here with probably what is a very simple question for most. I hope I am posting this in the right place. The question is:

A ball is thrown vertically into the air, and leaves the thrower's hand when it is 1.6m above the ground. It hits the ground 3.1s later.

Assume acceleration as 9.8, and assume no air resistance.

You have to find out everything really - initial velociy, maximum hieght, and speed at impact.

It seems quite simple but I can't get anyway.

I tried assuming displacement as 1.6m and a as 9.8, and did 1.6m=3.1t+4.9t^2. I got the answer 14.7m/s but you can't do that can you as a changes from + to -. Our teacher reckons the answer is closer to 12 but isn't really sure.

I also tried some simultaneous questions but ended up with stuff like 4.9t^2=4.9t^2

So what is the correct method?

Thanks very much

 

[Hootenanny]

Welcome to the Forums,

Your almost there with your first method, note that the ball finishes lower than it's initial position. What does this tell you about the sign of the displacement? Also note that the correct formula is;

\Delta x = v_{i}t + \frac{1}{2}at^{2}

It may have been a typo in your above post but your equation seemed incorrect.

 

[TicTac2]

Welcome to the Forums,

Your almost there with your first method, note that the ball finishes lower than it's initial position. What does this tell you about the sign of the displacement? Also note that the correct formula is;

\Delta x = v_{i}t + \frac{1}{2}at^{2}

It may have been a typo in your above post but your equation seemed incorrect.

I think I did have the equation correct I just shortened 1/2xa to 4.9.

The negative displacement is the mistake I'm making.

Thanks very much for the info!

 

[Hootenanny]

I tried assuming displacement as 1.6m and a as 9.8, and did 1.6m=3.1t+4.9t^2.

This should be v_it or 3.1v_i as 3.1 is the time t. A minor typing error though as it looks like you worked it out correctly.

My pleasure