A parachustist jumps from an airplane and freely falls y=51.8 m before opening his parachute. Thereafter, he decelerates at a=1.86 m/s^2. As he reaches the ground, his speed is 3.04 m/s.

a)How long was the parachutist in the air?

b)At what height did the parachutist jump from his plane?

There are two parts to the motion:

1. free fall over a distance of 51.8 m.
initial velocity, u=0 m/s
final velocity, v to be determined.
We ignore air resistance for this part.
acceleration due to gravity, a = -g = -9.8 m/s/s
Using the formula
v²-u² = 2aS
v can be determined.
(see answer to next question if necessary).
Time t1 = (v-u)/a

2. Parachute open (in infinitesimal time).
initial velocity, v (determined from part 1) m/s
final velocity,w = 3.04 m/s
acceleration, a = -1.86 m/s/s
Time, t2 = (w-v)/a
The distance descended when parachute is open can be determined also by:
w²-v² = 2aS

Total time = t1+t2

Add up distance for parts 1 and 2 will give the total distance.

To solve the given problem, we can use the equations of motion for uniformly accelerated motion.

a) To find how long the parachutist was in the air, we can use the equation:

v = u + at

Where:
v = final velocity = 3.04 m/s
u = initial velocity (when the parachute opens, the velocity becomes 0)
a = acceleration = -1.86 m/s² (negative because the parachutist is decelerating)

Rearranging the equation, we have:

t = (v - u) / a

Substituting the given values, we get:

t = (3.04 - 0) / -1.86

Simplifying:

t = -3.04 / -1.86

t ≈ 1.64 seconds

Therefore, the parachutist was in the air for approximately 1.64 seconds.

b) To find the height from which the parachutist jumped, we need to calculate the initial velocity (u) when he jumped from the plane.

Using the equation:

v² = u² + 2as

Where:
v = final velocity = 0 m/s (when the parachute opens)
u = initial velocity (when the parachutist jumps from the plane)
a = acceleration = -9.8 m/s² (acceleration due to gravity)
s = displacement = -51.8 m (negative because the parachutist is falling downwards)

Rearranging the equation, we have:

u = √(v² - 2as)

Substituting the given values, we get:

u = √(0 - 2 * -9.8 * -51.8)

Simplifying:

u = √(0 - 1013.68)

u ≈ √(1013.68)

u ≈ 31.85 m/s

Therefore, the parachutist jumped from a height of approximately 31.85 meters.