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

To find the height h, we can use the equation of motion for an object in free fall:

h = ut + (1/2)gt^2

Where:
h is the height
u is the initial velocity (which is zero in this case, as the body is freely falling)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time taken by the body to fall

We are given that the body describes 11h/36 in the last second of its fall. Let's analyze this information:

In the last second of its fall, the time, t, is 1 second.
The distance described, 11h/36, is equal to the height h.
So we can substitute these values into the equation:

h = 0 + (1/2) * 9.8 * (1)^2

Simplifying the equation gives us:

h = 0 + 4.9

Therefore, the height h is 4.9 meters.

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

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

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