Finding the acceleration of a soldier dropeed into a snow bank

[leprofece]

During the second world war, the Russians left his troops on the banks of snow falling from airplanes
low speed and flight flush, without the use of parachutes. Suppose that a soldier is dropped from an airplane traveling horizontally at a speed of 72 km/h at an altitude of 20m above a deep snow Bank. The soldier sinks to a depth of 2, 40m in the snow before stopping. What is the magnitude of the acceleration of the soldier while holding his movement in the snow?
Biook Answer a = 117; 85 m/s.

A FRIEND ANSWER:
you have to take into account that the soldier is released from 20 meters, so just before reaching the ground we can calculate its speed using conservation of mechanical energy: initial potential energy while falling soldier, you are turning into kinetic energy. EP = Ec--> m * g * h = 1/2 * v ² * m--> v ² = 2 * g * h = 2.10 * 20 = 400--> v = 20 m/s
Now we know that it has traveled a distance vertically from 2.4 m, so Vf'² - Vi ^ 2 = 2 * to * s--> 0 - 20 ^ 2 = 2 *(-10*2.4-> a = 83.33 m/s^2)
Note that extra information, the horizontal speed, that has no influence in what we know
To him the speed was not used in his solving

Who has the reason?

 

[HallsofIvy]

Re: Who stands the reason?

During the second world war, the Russians left his troops on the banks of snow falling from airplanes
low speed and flight flush, without the use of parachutes. Suppose that a soldier is dropped from an airplane traveling horizontally at a speed of 72 km/h at an altitude of 20m above a deep snow Bank. The soldier sinks to a depth of 2, 40m in the snow before stopping. What is the magnitude of the acceleration of the soldier while holding his movement in the snow?
Biook Answer a = 117; 85 m/s.

A FRIEND ANSWER:
you have to take into account that the soldier is released from 20 meters, so just before reaching the ground we can calculate its speed using conservation of mechanical energy: initial potential energy while falling soldier, you are turning into kinetic energy. EP = Ec--> m * g * h = 1/2 * v ² * m--> v ² = 2 * g * h = 2.10 * 20 = 400--> v = 20 m/s
Now we know that it has traveled a distance vertically from 2.4 m, so Vf'² - Vi ^ 2 = 2 * to * s--> 0 - 20 ^ 2 = 2 *(-10*2.4-> a = 83.33 m/s^2)
Note that extra information, the horizontal speed, that has no influence in what we know
To him the speed was not used in his solving

Who has the reason?

You can use the given depth, 2.4 m, to determine the time it takes to stop. But the acceleration is a vector quantity. You can determine the horizontal component of acceleration from the time to come to a stop.