Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9

That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor
It would be 22 depths inside the glass of water.
We have,
- Jane mass is 55 kg
- His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000
We know that,
Let's calculate force as we already have area;
- F = ma
- F = 55 × 9.8 { Acceleration due to gravity }
- F = 539 N
Now, if should she would be on 700 nails then pressure will be;
- P = F/A
- P = 539/7 × 10000
- P = 5390000/7
- P = 770,000 Pascal
And if should would be on a 1 nail only,
- P = F/A
- P = 539/1 × 1000000
- P = 539000000 Pascal
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It stops accelerating when the air resistance is equal to its weight.
That's (m•g)
= (2 kg) • (9.8 m/s^2)
= 19.6 newtons