Question

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.

Answer 1

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