Aparachutist falls 50.0 m without friction. when the parachute opens, he slows down at a rate of 61 m/s^2. if he reaches the ground with a speed of 17 m/s, how long was he in the air (in seconds)?

55.66 m

Explanation:

While falling by 50 m , initial velocity u = 0

final velocity = v , height h = 50 , acceleration g = 9.8

v² = u² + 2gh

= 0 + 2 x 9.8 x 50

v = 31.3 m /s

After that deceleration comes into effect

In this case final velocity v = 17 m/s

initial velocity u = 31.3 m/s

acceleration a = - 61 m/s²

distance traveled h = ?

v² = u² + 2gh

(17)² = (31.3)² - 2x 61xh

h = 690.69 / 2 x 61

= 5.66 m

Total height during which he was in air

= 50 + 5.66

= 55.66 m

newton's first law of motion