Question

A parachutist bails out and freely falls 50 m. Then the parachute opens, and thereafter she deceler- ates at 2.0 m/s2. She reaches the ground with a speed of 3.0 m/s. (a) How long is the parachutist in the air? (b) At what height does the fall begin?

Answer 1

Answer:

a)The parachutist in the air for 12.63 seconds.

b)The parachutist falls from a height of 293 meter.

Explanation:

Vertical motion of  parachutist:

 Initial speed, u = 0m/s

 Acceleration, a = 9.81 m/s²

 Displacement, s = 50 m

 We have equation of motion, v² = u² + 2as

 Substituting

   v² = 0² + 2 x 9.81 x 50

    v = 31.32 m/s

 Time taken for this

          31.32 = 0 + 9.81 x t

            t = 3.19 s

 After 50m we have

 Initial speed, u = 31.32m/s

 Acceleration, a = -2 m/s²

 Final speed , v = 3 m/s

 We have equation of motion, v² = u² + 2as

 Substituting

   3² = 31.32² - 2 x 2 x s

    s = 243 m

 Time taken for this

          3 = 31.32 - 2 x t

            t = 9.44 s

a) Total time = 3.19 +  9.44 = 12.63 s

    The parachutist in the air for 12.63 seconds.

b) Total height = 50 + 243 = 293 m

    The parachutist falls from a height of 293 meter.