Question

. A helicopter starts from rest t = 0 from ground with a certain constant acceleration in upward direction. Att=1s, a ball is
gently released from the helicopter and at t = 5 s, its engine is switched OFF. The ball strikes the ground at 1 = 3 s.
(i) Find the acceleration of the helicopter for 0<t<5 s. (ii) Find the height of helicopter from ground at t = 5 s.
(iii). Find the total time of flight of helicopter.​

Answer 1

Answer:

8 m/s²

100 m

15 sec

Explanation:

A helicopter starts from rest t = 0 from ground with a certain constant acceleration in upward direction

Let say Acceleration of Helicopter  =  H m/s²

Velocity at 1 sec = V = U + at = 0 + H* 1 = H m/s

Distance Covered = Ut + (1/2)at² = 0 + (1/2)H*1² = H/2 m

Velocity of Ball Dropped = H m/s upward

Distance to be covered = - H/2 m  in 2 sec  ( 1 to  3 sec strikes to ground)

H*2 - (1/2)*10*2² = - H/2

=> 2H - 20 = -H/2

=> H = 8

acceleration of the helicopter = 8 m/s²

the height of helicopter from ground at t = 5 s

= (1/2)H*5² = (1/2)* 8 * 25  = 100 m

at 100 m  , 5 sec engine is switched off

Velocity of Helicopter at 5 Sec = 0 + 8*5 = 40 m/s

Time taken further to reach ground

40 * t  - (1/2)*10t² =  - 100

=>  5t² - 40t - 100 = 0

=> t² - 8t - 20 = 0

=> t² - 10t + 2t - 20 = 0

=> (t - 10)(t + 2) =0

t = 10 sec

total time of flight of helicopter.​ = 5 + 10 = 15 sec