A body freely falling from a height h describes 11h/36 in the last second of its fall. Find the height h?

Answer 1

the position falling from a height of h feet is

y = h - 16t^2
After t seconds, y=0, so
h=16t^2

So, we have

(h-16t^2) - (h-16(t-1)^2) = 11h/36
0 - (16t^2 - 16(t-1)^2) = 11(16t^2)/36
32t-16 = 44/9 t^2
11t^2 - 72t - 36 = 0
t = 7.01

So, h = 16t^2 = 786.24 ft

Answer 2

I may have messed up the sign change, so double check my math.

Answer 3

the equations of a motions for a body falling down freely under gravity