Question

A parachutist bails out and freely falls 69 m. Then the parachute opens, and thereafter she decelerates at 2.6 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Height through which the parachutist free falls = H₁ = 69 m

Initial speed of the parachutist = V₁ = 0 m/s

Speed of the parachutist when she opens the parachute = V₂

V₂² = V₁² + 2gH₁

V₂² = (0)² + 2(9.81)(69)

V₂ = 36.79 m/s

Time the parachutist free falls for = T₁

H₁ = V₁T₁ + gT₁²/2

69 = (0)T₁ + (9.81)T₁²/2

T₁ = 3.75 sec

Acceleration of the parachutist after she opens the parachute = a = -2.6 m/s² (Negative as it is deceleration)

Speed of the parachutist when she reaches the ground = V₃ = 3.3 m/s

Time period the parachutist falls after opening the parachute = T₂

V₃ = V₂ + aT₂

3.3 = 36.79 + (-2.6)T₂

T₂ = 12.88 sec

Distance the parachutist falls through after opening the parachute = H₂

V₃² = V₂² + 2aH₂

(3.3)² = (36.79)² + 2(-2.6)H₂

H₂ = 258.2 m

Total time the parachutist is in air = T

T = T₁ + T₂

T = 3.75 + 12.88

T = 16.63 sec

Height at which the fall begins = H

H = H₁ + H₂

H = 69 + 258.2

H = 327.2 m

a) Time period the parachutist is in the air = 16.63 sec

b) Height from which the fall begins = 327.2 m