A Skydiver falls freely from the height of 3000 m with an acceleration of 10ms-2 and proceeds on opening the parachute after 10s. Find total duration of her jump, If she hits the ground at a velocity of 2 m/s
Answers
Given : A Skydiver falls freely from the height of 3000 m
acceleration of 10m/s² and proceeds on opening the parachute after 10 sec
she hits the ground at a velocity of 2 m/s
To Find : total duration of her jump,
Solution:
V = u + at
Velocity after 10 Secs = 0 + 10 (10) = 100 m/s
S = ut + (1/2)at²
Distance covered in 10 secs = (1/2)10 *10² = 500m
Remaining Distance = 3000 - 500 = 2500
Final Velocity = 2 m/s
Using V² - U² = 2aS
=> 2² - 100² = 2a(2500)
=> a =- 1.9992 m/s
V = u + at
=> 2 = 100 - 1.9992 t
=> t = 49.02 sec
total duration of her jump = 10 + 49.02 = 59.02 secs
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