If there was no air drag, how fast would drops fall from a cloud 1 km above the earths surface?

Question

The answer is 141 m/s. The equation used is t=v2d/g. I have a quiz this evening, and I can't seem to figure out how they arrived at the answer using this equation. (the book does not show the equation in action). 5 points to the first person who can show me how this is worked.

Answer 1

distance=d

acceleration=a=9.8

d=ut+0.5a(t*t) {commonly known equation}

u=initial speed=0m/s

therefore,t=squareroot[2d/a]

t=sqrroot[2*1000/9.8]=14.28 seconds to fall to the grnd

Answer 2

You want to know how fast (ideally) a rain drop would be traveling after falling from a distance of 1 km above the Earth's surface.

To answer this question, we will use this formula,
v_f^2 = v_i^2 + 2ad
Where v_f and v_i are the final and initial speeds of the rain drop respectively, a is the constant acceleration the rain drop experiences (we will assume that the acceleration is constant and equal to g, 9.81 m/s^2), and d is the distance over which the rain drop accelerates.

Let us assume that the rain drop starts from rest at a distance of 1000 meters above the ground.
In this case,
v_i = 0 m/s
d= 1000 m
a = g= 9.81 m/s^2 towards the ground
We only need to plug in to find v_f^2

v_f^2 = 2(9.81 m/s^2)(1000 m)
v_f^2 = 19620 (m/s)^2

In order to find the final speed, take the square root of this value,
Final speed = sqrt (v_f^2) = sqrt (19620 m^2/s^2)
Final speed, v_f = 140.07 m/s

Answer 3

Your answer is in the unit of velocity, not time. Since velocity is displacement/time, then your equation needs to be rewritten to correctly get you where you want to be. Remember that gravity (g) is equal to 9.8 m/s/s. Also remember to use the correct units when dealing with distance. Your given distance in km, but your equation uses meters.

Good luck on that quiz!