Hello all, I'm a 15 year old student in year 11 in Tasmania, Australia. This was a question from my Physical Sciences class that I was a bit confused about. Any help would be greatly appreciated. 1. The problem statement, all variables and given/known data A stone is thrown vertically into the air and caught at the same height. It leaves the thrower's hand at 29.4ms^-1. 1. Find the time taken for the stone to reach its maximum height. 2. The maximum height reached. 3. The time taken for the stone to return to the height from which it was thrown. 4. The stone's velocity as it hits the thrower's hand on its return. 2. Relevant equations I have figured out so far that: S=0 u=29.4ms^-1 a=-9.81ms^-1 3. The attempt at a solution 1. Is this just 29.4/9.81 ? 2. The above answer x 9.81? 3. Answer number one x2? 4. Would you use v^2=u^2 + 2as ? Thanks in advance.