- #1

TicTac2

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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