Thanks for the catch! I have the square in the algebraic equation, and the square was put into the calculator, but you're correct, I forgot to write it when I plugged in the numbers! Like I said, it was put into the calculator correctly, and I verified the answer, so it is correct at the end. Thanks for the comment!
Great question! If we were to draw a coordinate grid, regardless of where you put y=0 (you could put it at the climber or you could also put it at the water, anything moving down towards the water will be moving in the -y direction. Velocity is magnitude AND direction of course, so the initial velocity in the y direction is -2! Does that help? 1:27
@@MasteringSolutions oh I thought questions already took account of the displacement when they give the initial velocity. Thanks for helping me clarify
will never be able to thank bro enough
You forgot to square time at 7:30 when plugging into the 1/2at^2 portion of the equation
Thanks for the catch! I have the square in the algebraic equation, and the square was put into the calculator, but you're correct, I forgot to write it when I plugged in the numbers! Like I said, it was put into the calculator correctly, and I verified the answer, so it is correct at the end. Thanks for the comment!
why is initial velocity(vi) -2 instead of 2 ?
Great question! If we were to draw a coordinate grid, regardless of where you put y=0 (you could put it at the climber or you could also put it at the water, anything moving down towards the water will be moving in the -y direction. Velocity is magnitude AND direction of course, so the initial velocity in the y direction is -2! Does that help? 1:27
@@MasteringSolutions oh I thought questions already took account of the displacement when they give the initial velocity. Thanks for helping me clarify