Grade 11 math functions

If you use    d= 4.9+ 9.8 (t-1)   as the distance you fall EACH second  you will have to ADD the distance fallen each second to get the total meters fallen:

    4.9 + 9.8(8-1) +                         (73.5)

    4.9  + 9.8(9-1) +                         (83.3)

    4.9 + 9.8(10-1)  +                        (93.1)

    4.9 + 9.8 (11-1) +                        (102.9)

    4.9 + 9.8(12-1) +                          (112.7)

       TOTAL =   =============>  465.5     m    as before      Yah?

Let's first find your velocity at 7 seconds   ....we will assume there is no air resistance (you will not reach terminal velocity)

v = vo + at       vo = 0   t = 7     v then equals  9.8(7) = 68.6 m/s

now how far will you fall in the next 5 seconds (I will assume you want 5 seconds)

x = xo + vo t + 1/2 at^2

   = 0   + 68.6(5) + 1/2(9.8)(5^2) = 465.5 m  ====> distance you will fall from end of 7th sec to end of 12th second

First make a table 

1 4.9

2 14.7

3 24.5

Now find the pattern

To do that, subtract the second one from the first one

14.7-4.9=9.8

Check if it works for the second one

14.7+9.8=24.5

It does now use it

there are 5 seconds between 7 and 12 so 5*9.8=49

49 is your answer!!

(trust ElectricPavlov more than me)

This could be modeled by a sraight line function [even though it's not realistic ]

D  = 4.9 + 9.8 ( t - 1)  where D is the distance fallen in t seconds

So......between 7 and 12 seconds we have.....

D = [ 4.9 + 9.8 (12 - 1)]  - [ 4.9 + 9.8 (7- 1)]  = 49 ft  [ as gobromash found ]

Note -  EP's answer is "real-world"

Ah...I see what you mean, EP....I was taking it as the cumulative distance........4.9 ft in one second .....14.9 ft  total after two seconds, etc......I think your interpretation is correct.....