A 90kg man dives directly off of a 12m tower. What is his velovity when he strikes the water?
917853
I do not think this man has 90 g brain in his head doing such a stupidity.
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
8909
No, because the pressure at the surface will be less than 1 atm (or 1013 hPa). Pressure drops with altitude at a 100 hPa / 850 m ratio. Hence, if you dive at 1000 mt above sea level, pressure would be 778 hPa, = 0,77 atm at surface instead of 1 atm. Most diving computers have specific settings to calculate bottom times when diving at high altitudes. Check out a diving manual.
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
835975
This question can be answered with both calculus and standard constant acceleration equations.  In constant accelaration, velocity equals acceleration multiplied by time (v = at). Taking the integral of this gives us popular phyics equation d = v0t + (1/2)at^2, or distance equals initial velocity multiplied by time plus half the acceleration multiplied by the time squared.  Using these equations we have:\n\nAt peak of the dive:\nv = v0 - at\n0 = 12 - 9.8t\n9.8t = 12\nt = 1.22 (time in the air at peak)\n\nHeight at peak:\nd = vot + (1/2)at^2\nd = 12(1.22) + (1/2)(-9.8)(1.22)^2\nd = 7.34684 feet\n7.34684+10 feet (height of board) = 17.34684 feet above water\nSo, the answer to A is 17.347 feet\n\nWhen she will hit water:\nd = (1/2)at^2\n17.34684 = (1/2)(9.8)(t^2)\nt^2 = 3.54017\nt = 1.8815 (falling) + 1.22 (to reach peak) = 3.1015 seconds\nSo B is 3.1 seconds\n\nShe was falling for 1.8815 seconds at 9.8 m/s, so using v = at we have:\nv = 9.8(1.8815)\nv = 18.4387 m/s\nSo C is 18.4387 meters per second.
824337
Assuming that the boat is not moving under constant power, and neglecting the force of water resistence, the boat's new velocity would be 9.33 m/s.  \n\nWe know this because of conservation of momentum.  The momentum after the collision must be the same as the momentum before the collision.  Since momentum is equal to mass * velocity, the boat's momentum before the collision is:\n\n510kg * 11m/s = 5610kg*m/s. \n\nSince the momentum is the same after the collision, we just divide the total momentum by the new combined weight to get the new velocity:\n\nV = 5610kg*m/s / (510kg + 91kg) = 9.33m/s  \n\nThe speed with which Batman hit the boat does not factor into the equation, unless we're interested in the boat's angular velocity.  I'm assuming that you only want to know the boat's horizontal velocity over the water.  Since Batman had no horizontal momentum (he was falling straight down), his movement would not affect the horizontal motion of the boat.
502100
No ... I new a life guard who could not have been more than 90 lb... It was kinda funny to see her practising to save a 300 lb man though.
1063089
It depends on your age and build but average should be 59.1 kgs which is 130lbs. It could also range anywhere from 53.2 to 65 kgs ( 117 to 143lbs).
569938
sure!  you'll have a life vest on.  and the whole point is NOT to fall in the water, lol.  it's great fun!
9545
This is one of the questions that will always remain unanswered. But the answer should be that the other divers will try to save him. The life of a person is more important that a prize in a competition.
414287
Probably could, but there is a small chance you'll get bent doing it.  Is it worth the risk?  Keep this in mind:  Helicopters, when transporting divers, cruise at a ceiling of 50'.  That is as high as they're willing to go.  \n\nChanges in altitude are very dangerous after diving and should be done cautiously.
566593
100 m above sea level
517312
300 meters (or 1000 feet) is about as far as a human being can go without the aid of a vessel. Dive depth records are a little over 300 m.\n\nThe Eiffel Tower is 300 meters tall, which is not very far at all considering how deep the ocean goes. The ruins of the Titanic, for example, are at a depth of 3,800 meters (which is about 2.5 miles). The pressure at that depth is over three tons per square inch - enough to crush you in an instant!
299081
START SLOW HITTIN THE WATER OFF A BOARD AT HIGH SPEED CAN BREAK YA
1149881
No. You'd fall through it. It's just water molecules. It sounds cool though!
742135
There are several forces at work here.\n\nHorizontal: drag due to air friction slowing the horizontal velocity down\nVertical: weight, speeding the vertical velocity up\n\nThe total velocity when breaking surface tension is the vector sum of the horizontal (vh) and vertical (vv) velocities = V = sqrt(vh^2 + vv^2).  The vertical velocity = vv = gt = g sqrt(2S/g) = sqrt(2Sg); where S = 60 ft and g = 32.2 ft/sec^2.\n\nWe can't find vh with what you've given us because the initial horizontal velocity vh0 = 150 mph will be slowed down by the drag friction.  And drag friction forces depend on air density, coefficient of drag, and the cross sectional area of the person...none of which has been given.\n\nHowever, one rule of thumb in such matters is that the terminal velocity of a person falling through the atmosphere is about 120 mph.  So we can suppose that vh1, the end horizontal velocity, will be somewhere between 120 and 150 mph.  Let's assume the soft landing and vh1 = 120 mph = 176 ft/sec as the terminal horizontal velocity.\n\nThus, the break surface tension velocity V = sqrt(vh^2 + vv^2) = sqrt(176^2 + (2Sg)^2) = sqrt(31000 + 3800) = 186 ft/sec or about 127 mph.  Ouch.  At those speeds, water tends to act like concrete...in other words, splat...end of XXX.