physics energy lost to air friction?

Question

A parachutist of mass 76.3 kg jumps out of an
airplane at a height of 1280 m and lands on
the ground with a speed of 4.16 m/s.
The acceleration of gravity is 9.8 m/s2 :
How much energy was lost to air friction
during this jump? Answer in units of kJ.

Answer 1

There are several ways to go about this. I think the most straightforward is to first calculate the velocity (and thus the kinetic energy) of the parachutists had there been no parachute and no air friction.

so first calculate the time it takes to fall that distance with no air resistance, x= 1/2 a t^2 (no initial downward velocity)

1280 m = 1/2 (9,8) m/sec^2 t^2
261.2 sec^2 = t^2
16.2=t

then calculate the velocity achieved falling that long with no air resistance

v=at
v=9.8 m/sec^2 * 16.2 sec
v=158.4 m/sec

so, the kinetic energy of the paratrooper if he fell without resistance would have been

k= 1/2 m v^2
k=.5*(76.3 kg)*(158.4 m/sec)^2
k= .5 * (76.3 kg ) * 25088 m^2/sec^2
k=957107 kg-m^2/sec^2

knowing that 1 kJ=1000 kg-m^2/sec^2 then

k=957 kJ

Now we know the kinetic energy of the parachutist in the case of a true free fall

the kinetic energy for the given soft landing would be

k=1/2 (76.3) kg (4.16 m/s)^2
k=.5*76.3*17.3/1000=

Then the difference between the two energies is the loss.

Good luck.

Answer 2

Initial potential energy = mgh = 76.3 x 9.8 x 1280 = 957.1Kj
Kinetic energy on hitting the ground = 1/2mv^2
= 1/2 x 76.3 x 4.16^2 = 0.66Kj

Energy lost = 957.1 - 0.66 = 956.4Kj (Almost all of it)

Answer 3

what's their ability capability on the initiating of the bounce? Mass x height x gravity. you may artwork that bit out. what's the kinetic capability at touch down? ½mv² - ½ x 50 x 25 = 625 Joules. Subtract the final discern from the single you worked out for the aptitude capability - wager the place the version went.