pls help!!! please...challenge problem?

Question

challenge problem:
a raindrop falls from acloud 100m above the ground. neglecting air resistance, what is the speed of the raindrop when it hits the ground?



please with solutions...i need this to pass in my physics subject

our teacher said we should formulate our own formula...

our teacher said that we should write it completely if we forgot to put / she will disqualify us....

Answer 1

That depends upon the raindrop's radius. You CANNOT neglect air resistance. Tell your teacher that !

Just in case you could neglect it: s = 1/2g t^2 ==> t = sqrt (2s/g )
now v = gt ==> v = g . sqrt ( 2s / g )

with s = distance ( 100 m )
v = terminal velocity
g = the Earth's gravitational acceleration: 9.81 m / s^s
t = the time for the raindrop to reach the ground

so here v = 9.81 . sqrt( 2 * 100 / 9.81 ) or about 44 meters / second, which is about 100 miles / hour

Engineer

Answer 2

I see that ypur teacher wants you to solve this using first principles.
As defined in the problem, let us negelct the air resistance.
When a rain drop (or any object for that matter) falls from a height, it is acclerated to the earth surface uniformly. What this means is that the velocity increases from zero at 100m to a "V" just before it hits the ground. The increase in velocity is the same for a give time interval. In other words it is attracted by earth at a rate of 'g' (9.8 m/sec2)

At any point of time the velocity can be represented as

v = u + g * t

v = instantaneous velocity at any time "t"
where u = initial velocity.

In this case initial velocity is "zero" to start with.

So we have v = g * t

Assume it takes "T" seconds to reach the ground.
Then we have Velocity when it hits the ground "V" = 9.8 * T

Now since the velocity uniformly increases from "zero' to "V" at the ground, the average velocity is V/2.

Total time taken = Distance travelled/Average velocity
T = 100/(V/2) = 200/V

So V = 9.8 * 200/V
ie V2 = 1960

Therefore V = 44.271m/sec

Alternate approach
------------------------

When a object falls from height "h", its potential energy is converted to kinetic energy. Assuming no energy lossses (we are neglecting air resistance), we have,

PE = KE
Potential energy = m*g*h
Kinetic energy = 0.5 * m*v*v

equating both these,

mgh = 0.5 *m*v*v

g*h = 0.5 *v*v
v = sqrt(2gh) = sqrt(2*9.8*100) = 44.271 m/sec

Answer 3

Energy Potential=Energy Kinetics

m g h = 1/2 m v^2
g h = 1/2 v ^2 h=100 meters
2 g h = V^2 g=9.81 meter/sec^2
2 9.81 100 = v^2
V = sq.root 1962 m/sec

Answer 4

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Answer 5

You need to use the formula (vf)^2 = (vo)^2 + 2ax. vf is the final velocity, vo is the initial velocity (zero in this problem), a is the acceleration (g = 9.8 m/s^2 in this problem), and x is the distance (y = 100 m in this problem). So rewrite the formula as (vf)^2 = 0 + 2gy ==> vf = sqrt(2gy) = sqrt(2*9.8*100) = 44 m/s.

Answer 6

It is quite an easy question...
Just use
v^2 = u^2 + 2as
{u=0
s=-100m
a=-10}

v=41(approx.)

Wasn't that easy...

Answer 7

potential energy lost falling = kinetic energy at bottom

mgh = 1/2mv^2

v = sqrt (2gh) = sqrt (2.10.100) = sqrt (2000) = 44.72 m/s

Answer 8

h = 1/2 g t^2, so you obtain t = square root ( 2*h/g)

Now

v = v_0 + g * t = 0 + g * t = square root (2*g*h)=square root(2*100*9.8) = 44.27 m / seg