Answer:
The deceleration of the pilot was 1.9 x 10⁴ m/s²
Explanation:
First, let´s calculate the velocity of the pilot 3 m above the ground. To do that, we need to know how much time the pilot was falling. The equation for the position and velocity of the pilot are as follows:
y = y0 + v0 * t + 1/2 * g * t²
v = v0 + g * t
where
y = position of the pilot at time t
y0 = initial position
v0 = initial velocity
t = time
v = velocity at time t
g = acceleration due to gravity
If we consider the ground as the center of our reference system and that the pilot fell with an initial velocity of 0 (the pilot would unlikely impulse himself to the ground), then the equations would be as follows:
y = y0 + 1/2 g * t²
v = g * t
The time at which the pilot was 3.0 m above the ground will be:
3.0 m = 6000 m - 1/2 * 9.8 m/s² * t²
3.0 m - 6000 m / -4.9 m/s² = t²
t = 35.0 s
The velocity at that time will be:
v = -9.8 m/s² * 35.0 s = -343 m/s
After 35.0 s the pilot has a positive acceleration besides the acceleration due to gravity. Then, the equation for velocity and position will be:
v = v0 + a*t now, v0 = -343 m/s and a ≠ g and a>0
y = y0 + v0 * t + 1/2 * a * t² now, y0 = 3 m
Again, let´s find the time at which the pilot hits the ground:
v = v0 + a*t
v-v0/ t = a
Replacing in the equation for position:
y = y0 + v0 * t + 1/2 * ((v-v0)/t) * t²
y = y0 + v0 * t + 1/2 * v* t - 1/2 * v0 * t)
y = y0 + 1/2 v0 * t + 1/2 v * t
replacing with numbers:
0m = 3m + 1/2 * (-343 m/s)t + 1/2 *(-54 m/s)t
-3.0 m = - 198.5 m/s * t
t = -3m / -198.5 m/s
t =0.015 s
the upward acceleration was then:
v-v0/ t = a
-54 m/s -(-343 m/s) / 0.015 s = a
a = 1.9 x 10⁴ m/s²
