. 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.
Answers
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