ryan g

Ok I think ive got it! many many thanks for all your help! Merged post follows: Consecutive posts mergedactually I dont got it! the constant i worked out to be 2.38 how do i then apply that number to work out the height? or is that the height? Merged post follows: Consecutive posts mergedhang on thats totally wrong - i used the wrong values. im getting really confused which equations to use - Im not getting taught kinematics. right start from the top. im covered in scraps of paper with all sorts of equations on an am totally lost.

ok - I plotted a rough graph with velocity against time - deducting 2.8 metres for each second. the time at 0 velocity is roughly 3.5 secs. Is that correct? Merged post follows: Consecutive posts mergedthis is so frustrating - I have all the rest of the answer planned - to get the final speed on impact with the ground - just finding the height and I know its so simple!! argh. I am now using the actual values stated in the question to work it out and the time the ball takes to reach speed 0 m s = 3.9 s (using initial speed of 7.8 m s rather than 7 m s and acceleration due to gravity -9.8 m s ball leaves hand at 1.4m above thr ground.) slotting this in the equation you gave me gives me x = -42.68m !!!!!! surely impossible

aha - im getting it now maybe trying to cram the whole book on energy and light in 2 days isn't such a good idea! I have that equation written down somewhere - I'll have a go in the morning and let you know how I get on many thanks for your time and patience!!

sorry - i don't know eithier of those equations and have no idea how to adapt them to get the height. is 2.5 secs correct? so distance travelled = speed x time taken = 7 m s x 2.5 s = 17.5 m then do I do this : 17.5 - 9.8 = 7.7 m 7.7 m + 1.6m = 9.3 m ?????????

7 / 2.8 = 2.5 2.5 secs so after leaving the hand the ball is in the air for 2.5 secs before reaching its maximum height. that seems reasonable - am i getting warmer?

I dont think this is right but v = change in speed / time taken 7 = - 9.8 / t x t t7 = -9.8 /7 t = 1.4 i dont think Ive got the right equation. is magnitude of acceleration the same as velocity? really sorry - the answer is probably staring me in the face!

Ive learned the equations for kinetic and potential energy - no idea about the other. Im still a bit confused - how do I apply them to work out the hieght? sorry if this appears dumb - im returning to education / learning at home for the first time in ten years and its a bit of a nightmare trying to get my head around everything. thank you thank you Merged post follows: Consecutive posts mergedwould I be right in thinking as the ball is travelling at 7 m s when thrown and the gravitational energy is 9 m s the ball will stop travelling at 2 metres above the release point. or is that just wrong?

OK - what if I wanted to find the speed that the ball is travelling when it hits the floor? The question I need to answer simply requires the speed on impact with the ground. Im a bit worried about working out how long the ball will take to reach its maximum hieght because Im not sure ive been taught how to do that yet and not sure if its necessary for my answer Is it possible to work out the speed the ball is travelling when it hits the ground WITHOUT knowing the maximum height the ball reaches in the air? I have worked out an answer but the speed is lower than that of which the ball was travelling when it was thrown. Im guessing thats incorrect due to acceleration due to gravity? any help is greatly apprieciated !

Hi first post on here - hope someone can help If I were to throw a ball in the air and the ball left my hand at: a speed of 7 m s -1 and at a height of 1.6 m above the ground how high would the ball go? what equation should be used to work this out? assuming the acceleration due to gravity is 9.8 m s -2 any help would be grand many thanks R