struggling with a level mechanics

A ball is thrown vertically upwards with a speed 0.9m/s from a point P at height h metres above the ground. The ball hits the ground 0.75s later. The speed of the ball immediately before it hits the ground is 6.45m/s. The ball is modelled as a particle. Find the height above P to which the ball rises before it starts to fall towards the ground again.

Reply 1

mqb2766

Original post by cool11111111111

Sounds like a typical suvat problem. Have you sketched it and listed the variables and which equations you could use?

Reply 2

cool11111111111

^{Yep. I've tried using v^2=u^2+2as, but my results keep coming out either as negative or really small numbers which don't make sense}

Reply 3

mosaurlodon

v^2=u^2+2as, set up as positive and down as negative, so u=0.9, v=0, a=-g.

The answer is a little small if thats why youre confused.

Reply 4

mqb2766

Original post by cool11111111111

^{Yep. I've tried using v^2=u^2+2as, but my results keep coming out either as negative or really small numbers which don't make sense}

Thats the right suvat, but to "understand" it you could note that
v = u + at
so the upward phase is only ~0.1s and the following downward phase is ~0.65s, so its not going to move that far in that time.