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:
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
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)
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 x t + 1/2 x 9.8 m s⁻² x 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, then from B to C is = t = t₁ + t₂ = 2 s + 6 s = 8 s --------------Ans.
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