Total distance fallen by end of nth second if d distance falls in nth second

[jemerlia]

Homework Statement

A body falls vertically from rest. During the nth second it falls a distance d. Prove that by the end of the nth second it has fallen a total distance (D) of (2d+g)^2/8g

Homework Equations

x = x0 + V0 + 1/2at^2
where
x0 = initial position
v0 = initial velocity
a = acceleration
t = time

The Attempt at a Solution

The total distance is D = 1/2gn^2 because the object has fallen for n seconds. At the end of the first second d=D and t^2 = 2d/g. I'm unsure about how to proceed from here. Any help gratefully received...

 

[PhanthomJay]

jemerlia, welcome to PF!

Hint: What is the distance fallen after (n-1) seconds?

 

[jemerlia]

Thanks for the hint - clearly the distance d is given by
d = 1/2gn^2 - 1/2g(n-1)^2
which (according to my rusty maths) simplifies
= 1/2g(2n-1)

It appears that t^2 = 2d/g is useful here but substitution appears to give nonsense. I 've obviously missed or misunderstood something.

 

[rl.bhat]

{d = 1/2gn^2 - 1/2g(n-1)^2
which (according to my rusty maths) simplifies
= 1/2*g*(2n-1)}
This is right. Now find n in terms of d and g.
Now the total distance fallen in n seconds is D = 1/2*g*n^2. Substitute the value of n. You will get the required answer.

 

[jemerlia]

Many thanks for the help during the "holidays"...

Because d = 1/2 * g * (2n-1) then

n = (2*d/g + 1) / 2

= d/g + 1/2

However, when n is substituted into

D = 1/2*g*n^2

as

D = 1/2*g*(d/g + 1/2)^2

it does not produce the expected result. I guess I have a problem with the arithmetic somewhere... advice gratefully received...

 

[rl.bhat]

D = 1/2*g*(d/g + 1/2)^2

D = 1/2*g*(2d + g)^2*1/4g^2
= (2d + g )^2/8g

 

[jemerlia]

Thank you to everyone who gave their time to help me with this problem. It has served to identify the areas I must work on.