Question

A raindrop falls to the ground from a raincloud at an altitude of 3000 meters.

Answer 1

Question is missing, by a search, you can find that we want to know how long takes raindrop to reach the ground and the velocity that the raindrop has when it touches the ground.

The acceleration will be the acceleration of gravity, this is:

a = -9.8m/s^2

for the velocity, we integrate over time, and because there is no initial velocity, the constant of integration will be zero, so we have:

v = -9.8m/s^2*t

again, for the position we integrate again, now, the initial position is 3000m, so we have a constant of integration,

r = -4.9m/s^2*t^2 + 3000m

the raindrop will touch the ground when r(t) = 0, so we must solve the equation:

-4.9m/s^2*t^2 + 3000m = 0

for t:

t = √(3000m/4.9) s = 24.7 s

So the raindrop takes 24.7 seconds to reach the ground, if you want to know the velocity of the raindrop when it hits the ground, you need to replace that time in the velocity equation:

v(24.7s) = -9.8m/s^2*24.7s = 242.06 m/s

So the velocity at which the raindrop hits the ground is 242.06 meters per second.

Answer 2

24.7 s

Step-by-step explanation:

Question is appear to be missing. Found it on google:

"a) If there were no air resistance, how long would it take to fall?"

Solution:

We can solve the problem by using the following equation of motion for a uniform accelerated motion:

`d=ut+(1)/(2)at^2`

where

d = -3000 m is the displacement of the raindrop (negative because in the downward direction)

u = 0 is the initial velocity of the raindrop if we assume it starts from rest

t is the time taken for the fall

a = g = -9.8 m/s^2 is the acceleration of gravity

Solving the equation for t, we find:

`t=\sqrt{(2d)/(g)}=\sqrt{(2(-3000))/(-9.8)}=24.7 s`