1.
Whatever Pe he had up on that board relative to the water surface is all Ke when he hits the water.
Ke = (1/2) m v^2
= (1/2) (75) (7.7)^2 Joules
THAT is the Pe he had up on the board :)
2. NOW YOU ARE ON A ROLL
Pe = m g h = answer to #1 above :)
Please help me with this!! I am really confused about it...
1. A 75 kg. man steps off the end of a diving board, and hits the water at 7.7 m/s. Ignoring air resistance, what was the mans PE (potential energy) right before he stepped off the diving board?
*I know that the formula for potential is equal to its weight multiplied times its height. And I know that weight is mass times acceleration due to gravity, so then PE = mgh. I don't know how to use this information to find the PE since I don't know that height of the diving board...
2. How high is the diving board?
I don't know how to find the height of the diving board... since i don't the PE because if I knew the PE i could find the height by switching the original equation (PE = mgh) and make it into (h = PE/mg). I really need to know how to find the height and the potential energy in this problem!! Please explain these to me!?! Thank you!!
5 answers
got it ?
Sorry I had something to do! Thank you for explaining it!! Let me just see if i understand it completely...
Soo then the PE before he jumps off the diving board is 2223.375 Joules?? And the height of the diving board is 29.645 meter??
agree with 2223 Joules
m g h = 2223
75 * 9.81 * h = 2223
h = 3.02 meters
m g h = 2223
75 * 9.81 * h = 2223
h = 3.02 meters
1 answer