"8 seconds" is not a velocity.
so what you need to do is find the velocity that the person enters the water and then use the equation v sub final squared = v sub initial squared + 2*acceleration(final distance-initial distance). final velocity is zero, find the initial velocity yourself and use 2 as the final distance where the initial distance is 0, solve for acceleration. Easy way: the decelleration would have been twice that provided by gravity because the diver decellerated to zero in half the distance of the dive. The diver starts at zero and hits max velocity in 4 meters then goes from max velocity to zero in 2 meters. 2 x 9.81 m/s2 = ___
Air resistance increases and terminal velocity decreases when the parachute has opened.
Vf^2 = Vi^2 +2ad = 1.8m ^2 + 2(-9.8m/s^2)(3.0m) =62.04m/s Vf= sqrt(62.04m/s) =-7.9m/s
We'll assume you mean 2 seconds AFTER he's jumped from the plane. A good estimate would be 9.8m/s times 2 (9.8m/s/s times 2 seconds -- the standard value of the acceleration due to gravity). In two seconds, the velocity would be low enough that air resistance could probably be ignored for purposes of answering your homework. So try 19.6 m/s, unless your teacher gave your data on the air resistance of the sky diver.
250N
80 ft / sec ~ 55mph The above ignores air resistance which will not be significant at this speed.
A diver springs from the edge of the ocean with an initial upward velocity of 8 ft/s. How long will it take the diver to reach the water?
473.69 ft. exactly i measured it with my ruler
Assuming that acceleration dut to gravity is 32 ft/sec2 and that air resistance is insignificant, the answer is 2.5 seconds. To show work, at 8 ft/sec v = at = 32 t t = 1/4 second jumping up and v = 0 before free fall s = 1/2 at^2 = 16t^2 = 16x.25 x .25 = 1 foot so diver free falls 81 feet 81 = 1/2 at^2 so t = 2.25 sec 2.25 + .25 = 2.5 sec
so what you need to do is find the velocity that the person enters the water and then use the equation v sub final squared = v sub initial squared + 2*acceleration(final distance-initial distance). final velocity is zero, find the initial velocity yourself and use 2 as the final distance where the initial distance is 0, solve for acceleration. Easy way: the decelleration would have been twice that provided by gravity because the diver decellerated to zero in half the distance of the dive. The diver starts at zero and hits max velocity in 4 meters then goes from max velocity to zero in 2 meters. 2 x 9.81 m/s2 = ___
Terminal Velocity.
Terminal velocity
Terminal Velocity.
Any skydiver can fall at a constant velocity. This is called the, "Terminal velocity". Since we do not have the picture that you mention, nothing more can be assumed.
Air resistance increases and terminal velocity decreases when the parachute has opened.
Vf^2 = Vi^2 +2ad = 1.8m ^2 + 2(-9.8m/s^2)(3.0m) =62.04m/s Vf= sqrt(62.04m/s) =-7.9m/s
Anything, regardless of weight, falls at 32 feet per second squared, until it reaches terminal velocity. Assuming no air resistance we need SQRT(27000/32) or 29.05 seconds to splat. However, Wikipedia puts the terminal velocity of a sky diver at 180.45 feet/second. That speed would be reached in about 6 seconds. 27000/180.45 = 149.63 seconds or 2 and a half minutes. So you're looking at 2 minutes 33 seconds +/- 3 seconds.