freely falling bodies ?

Working formula is

Vf^2 - Vo^2 = 2gh

where

V = final velocity = velocity of ball at its max height = 0

Vo = initial velocity

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

h = maximum height attained by ball = 25 m (given)

Substituting values,

0 - Vo^2 = 2(-9.8)(25)

NOTE the negative sign attached to the acceleration due to gravity. It simply means that the ball loses its speed as it goes up.

And, solving for Vo,

Vo = 22.14 m/sec.

Let

s = height where the speed of the ball is half of its original value

and using the same formula above,

Vf^2 - (22.14)^2 = 2(-9.8)(s)

Since Vf = (1/2)(22.14) = 11.07, then the above simplifies to

(11.07)^2 - (22.14)^2 = -19.6(s)

and solving for "s"

s = 18.75 m.

Hope this helps.

We can find the launch speed from

v^2 - u^2 = 2as

At max height v = 0

-u^2 = -2g x 15

u = SQRT(50g)

Now v = (1/2)u = (1/2)SQRT(50g)

From v^2 - u^2 = 2as

s = (v^2 - u^2)/-2g = -[(3/4) x 50]/-2 = 18.75 m

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Yes. Figure forwards to get original velocity then backwards to get height

d = 0.5 AT^2

25 = 0.5 * 9.8 m/s^2 * T1^2 solve for T1

Then

V = AT1

V1 = 9.8 * T (from above) Solve for V1

V2 = V1/2 Solve for V2

V2 = AT

V2 = 9.8 * T2 solve for T2

d2 = 0.5 * A*T2^2 Solve for d2

d2 = 0.5*9.8*T2^2

equation v^2=u^2+2as preliminary velocity u=0, then v^2=2gh(a=g,s=h) at top h, permit v be the fee then v^2=2gh .........eq a million at some top h1,it somewhat is velocity 2v strengthen into doubled then (2v)^2=2gh1 4(v^2)=2gh1 by making use of eq a million 4(2gh)=2gh1 cancelling 2g we get h1 =4h