free fall help?

v^2-v0^2=2gh

v^2-18^2=2*12*10

v=23.7 m/s

v=v0+gt

t=(v-v0)/g

t=(23.7+18)/10 s

t=4.17s

there are two ways of answering this question, one is by using the kinematics formulas involving velocity, displacement and and acceleration and the other is using the conservation of energy.

Using the law of conservation of energy:

KEo+ PEo= KEf + PEf

KEo = mv^2/2

PEo = mgh

(Initial energy) = fianl enrgy

mv^2/2 + mgh = mv^2/2 + mgh

since mass is constant, then the mass m will be canceled leaving;

(v^2)/2 + gh = (v^2)/2 + gh

(Note: v and h on the left hand side of the equation is different to the v and h of the right hand side so they will not cancel )

Taking the bottom of the 12m hole as our reference position, then we have (it means that potential energy at the bottom of the pit should be zero)

also we know that at the top of the flight the velocity = 0

that is: Vtop = 0

from this we can compute for the maximum height of the stone.

Hmax = (Vo)^2/2g

g =9.8 m/s^2

Hmax = (18m/s)^2/19.6m/s^2

= 16.53 m

the potential energy at the bottom is zero right?

then, PE max = KE max

PE max is achieved at Hmax, while KE max is achieved at the bottom (just before stone hits the ground)

Then the overall height is 12 m + 16.53 m = 28.53 m

then v = sqrt(2gh)

h = total height (28.53 m)

V final = 23.64715 m/s

total time of flight is given by t = 2Vinitial/g

t = 4.246 s

hehe

u = 18m/s.

a = -9.8m/s^2

displacement = – 12m.

We have to find both t and v.

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It will be easier if we first calculate v.

v^2 = u^2 – 2as

v^2= 18^2 + 2*(– 9.81)*(– 12)

v = ±23.65m/s Since it is falling down

v = – 23.65m/s

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time = v-u /a = [ – 23.65 -18] / (– 9.81)

t = 4.25 s.

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