- #1
TicTac2
- 4
- 0
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
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