A helicopter is rising up from the ground with an acceleration of g m/s-. starting from rest. after rising a height h, it attains a velocity of v m/s. At this instant, a particle is now released from the helicopter. Take t = 0 at releasing time. calculate the time t when partied reaches to the ground.
Answers
We are given a situation where a helicopter rises up from the ground with initial velocity u = 0 m/s and a = g m/s² reaches a height h with velocity, v m/s.
At that time a particle is released from the helicopter, we take t = 0 as the releasing time.
So, We need to find the time t when particle would reach the ground.
We have,
- Acceleration = g
- Final velocity = v
Using the third equation of motion,
⇒ 2gh = v² - 0²
⇒ 2gh = v²
⇒ v = √2gh ...(1)
Which is the initial velocity of the particle. but we took velocity in the upwards direction positive, So the particle coming downwards would have negative velocity.
∴ u of particle = -√2gh
Let the time taken be t,
⇒ h = ut + 1/2gt²
⇒ h = -√2gh×t + gt²/2
⇒ h = (-2t√2gh + gt²) / 2
⇒ 2h = -2t√2gh + gt²
⇒ gt² - 2√2gh×t - 2h = 0
Now, It is a quadratic equation, Comparing it with the standard form of a quadratic equation, we have
- a = g
- b = -2√2gh
- c = -2h
∴ t = -(-2√2gh) ± √[ (2√2gh)² - 4×-2h×g] / 2g
⇒ t = {2√2gh ± √(8gh + 8gh) } / 2g
⇒ t = (2√2gh ± 4√gh ) / 2g
There are two values of time,
Let us find one by one:
Case 1 (Positive sign)
⇒ t = (2√2gh + 4√gh) / 2g
⇒ t = (2√2gh + 2√2 × √2gh) / 2g
⇒ t = 2√2gh (√2 + 1) / 2g
⇒ t = √(2h/g) (√2 + 1)
Case 2 (Negative Sign)
⇒ t = (2√2gh - 2√2×√2gh) / 2g
⇒ t = 2√2gh(-√2 + 1) / 2g
⇒ t = -√2h/g (√2 - 1)
In case 2, time is negative, hence It is neglected.
Hence, The time taken by the particle is √(2h/g) × (√2 + 1)
∴ Option 3 is correct.
Some Information :-
◉ Other equations of motion:
- 1st : v = u + at
- 2nd : s = ut + 1/2 at²
- 3rd : 2as = v² - u²
◉ In a motion where air friction is negligible.
We can find,
- Total distance travelled = u² / g
- Time of flight = 2u / g