A helicopter is ascending vertically at a speed of 19.6 m/s. When it is a height of 156.8 m above the ground, a stone is dropped. How long does the stone take to reach the ground? (Ans. 8.0 s)


Explanation:


Due to inertia the stone will continue to move up with the same velocity as the helicopter and will stop moving up after gaining some height due to the force of gravity. Afterward it will start free falling toward the ground. 

Given:


The initial Velocity of the stone is the same as the helicopter = vi = 19.6 m s⁻¹
Initial height h₁ = 156.8 m

To Find:

Total time of flight of stone = t = ?

Solution:


The journey from Point A to B: we have
initial velocity of stone = vi =19.6 m s⁻¹
The final velocity of stone = vf = 0 m s⁻¹ (stone comes at rest at highest point)
Acceleration due to gravity = g = -9.8 m s⁻² (-ve due to upwards motion)
To Find: 
Height A - B = h₁ = ? and time = t₁ = ?

using 3rd equation of motion 

2 g h₁ = vf² + vi²

2 (9.8 m s⁻²) h₁ = (0 m s⁻¹)² + (19.6 m s⁻¹)²

19.6 m s⁻² h₁ =  (19.6 m s⁻¹)²

Or

h₁ =  19.6 m

Now to find the time taken from point A to B, we have 1st equation of motion

vf = vi + a t₁  

by putting values

0 m s⁻¹ = 19.6 m s⁻¹ + (-9.8 m s⁻²) t₁ 

or

(9.8 m s⁻²) t₁ = 19.6 m s⁻¹

or

t₁ = `\frac {19.6 m s⁻¹}{9.8 m s⁻²}`

t₁ = 2 s

The journey from Point B to C: we have
initial velocity of stone = vi =0 m s⁻¹ (start falling from rest)
Acceleration due to gravity = g = 9.8 m s⁻² (+ve due to upwards motion)
height h₃ = h₂ + h₁ =  156.8 m + 19.6 m = 176.4 m

So using 2nd equation of motion

h₃ = vi t + 1/2 g t²

putting values

176.4 m = 0 m s⁻¹ ï½˜ t + 1/2 ï½˜ 9.8 m s⁻² ï½˜ t₂²

or 

t₂² =  `\frac {176.4 m ｘ 2}{9.8 m s⁻²}`

t₂² =  36 s²

by taking square root on both sides we get

t₂ = 6 s

 Hence the total time of flight from A to B, and then from B to C = t = t₁ + t₂  = 2 s + 6 s  = 8 s --------------Ans.




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