A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 27 m/s^2. If he reaches the ground with a speed of 14 m/s, how long was he in the air (in seconds) ?

3.83 s

Explanation:

Initially for 50 m A falls with acceleration equal to g ie 9.8 m / s². In this journey his initial velocity u = 0, a = 9.8, v = ?, t = ?

h = ut + 1 / 2 g t²

50 =.5 x 9.8x t²

t = 3.19 s

v = u + gt

= o + 9.8 x 3.19

= 31.26 m / s

For motion under deceleration

initial speed u = 31.26 m/s

Final speed v = 14 m/s

deceleration a = - 27 m/s²

v = u - at

14 = 31.26 - 27 t

t = 0.64 s

So total time in the air

= 3.19 +.64 = 3.83 s