xylia Posted March 31, 2014 Share Posted March 31, 2014 (edited) i have a school problem to do, but have no idea where to start of how! a person of mass m wants to dive from a 3.0 m board above the water, by jumping off the board rises to a further 0.6 m, at what speed does m enter the water? knowing that g is 9.8 m s-1 ( acceleration due to gravity) Edited March 31, 2014 by xylia Link to comment Share on other sites More sharing options...
Fuzzwood Posted March 31, 2014 Share Posted March 31, 2014 Which relevant equations do you know? Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 v= square root ( 2 g delta h) or kinetic energy or work done by gravity work done against gravity Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 (edited) Sure, you must know some equations of motion under constant acceleration. To start you off I will redescibe what happens with annotated questions. The person jumps (which way) off a diving board. When he reaches his max 0.6m above the board 1) What is his velocity at this point? 2) What is his total height above the water? 3) What distance does he travel to hit the water? 4) What his acceleration? 5) what is his final velocity? Can you do the problem now? Edited March 31, 2014 by studiot 1 Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 I'm trying to understand, but i rely can't get it. i should calculate the velocity first against gravity with the jump then calculate the velocity by gravity with the initial speed of the jump? am i on the right path? Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 Have you drawn yourself a diagram? Do you know the formula for the final velocity, given the initial velocity, acceleration and distance covered? Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 i have drawn a diagram, but there are too many formulas and i can't rely find the right one because I'm rely comfused Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 (edited) Well your diagram should look something like an upside down hockey stick over the diving board. Look carefully at this diagram and my list of questions in post 4. You should be able to answer the first four just by looking at the diagram. You only need one formula to obtain your answer (my question 5) from this information, not lots of them. Let us work through the firstr four and then think about 5. Edited March 31, 2014 by studiot Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 v= squared( 2 g delta h) Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 (edited) Posted Today, 04:42 PM v= squared( 2 g delta h) Is this your answer for my question1? Which direction is he travelling from the diving board to point A? Which direction is he travelling from the point A to the water? So what must happen to the velocity, instantaneously, at point A? Edited March 31, 2014 by studiot Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 point A must be 0 m s Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 Voila you have an initial velocity, u=0. Now can you answer the how far questions? Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 adding 3.0 m+ 0.6m giving the full distance till the water surface Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 Voila you have u=0, s=3.6, What is the acceleration during the whole of the downward journey from A to the water? And then we return to the formula I asked for in post6 Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 i do not know this equation, the only equation in my text book for speed is the one i wrote above Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 (edited) With no acceleration, that is constant velocity. Distance = velocity x time s=ut; initial velocity ( = u) = final velocity (=v) If there is constant acceleration ( I will use f for its symbol) then we have sevaral formulae, you should have met before reaching the formula you quote. v = u + ft s = ut + 0.5ft2 v2 = u2+2fs The first two contain time, which we do not know but the third is the equation I was asking for. Now the acceleration (f) = g, u = 0 and s = 3.6 So substituting we have your formula v2 = 0 + 2g (3.6) Edited March 31, 2014 by studiot Link to comment Share on other sites More sharing options...
xylia Posted March 31, 2014 Author Share Posted March 31, 2014 thank you so much for your help, just one last question if we start from a speed of 0 m s can't i just write my formula anyway? Link to comment Share on other sites More sharing options...
studiot Posted March 31, 2014 Share Posted March 31, 2014 Yes, of course you can use your formula. But not all future questions will be the same so I hope this thread has helped you understand more. In answering questions I find it a good idea to list all the information provided as we did here and then list the equations that might be relevant and see how they match. Go well in your future studies Link to comment Share on other sites More sharing options...
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