What was the upward component of the average acceleration vector of the ball? Physics question?

You throw a ball upward with an initial speed of 4.52 m/s. When it returns to your hand 0.92 s later it has the same speed in the downward direction (assuming air resistance can be ignored). What was the upward component of the average acceleration vector of the ball?|||4.52^2 = 2gh





4.52^2 = 20.4304





20.4304/(2 x 9.8) = h = 1.0424 m





1.0424 = 1/2gt^2





1.0424/4.9 = t^2 = 0.2127, sq-rt = t = 0.46122





0.46122 x 2 = 0.9224 secs ( just checking your time given was correct, it is!)





(v - u)/t = acc (a)





(0-4.52)/0.46 = - 9.826 m/s^2 (answer)|||HyperPhysics.info is an awesome tool for physics homework. A quick search for "constant acceleration" gives:





http://hyperphysics.info/?cx=partner-pub…





Click the top link, that will give you the governing equations. Then click the second link and scroll down to the built in calculator to check your work.