Thread: Falling Object in Straight Line

An object falling in a vertical line passes a window 3 m high in 1/6 seconds. Find the distance above the top of the window from which the object let fall.

No idea about this question at all.

I did
a=9.8
thus v^2 = 19.8x
at v=18 (at end of window)
x=16.5
thus origin above window = 13.5

However answer is 15.06

Lets take velocity of object at the window's starting level as "v"

time taken by object to cross the window= 1/6 second

So using Second equation of motion

v*(1/6) + (9.8*(1/6)^2)/2 = 3

This gives v

Now Let us take the distance of origin to window = h

Initial velocity is 0

Velocity after traveling h distance is given by

Using Third equation of motion


v^2 -0^2 = 2*9.8*h

Thus you get h after putting v

Here's a great lecture that will explain how to do any problem of this type with no fuss, just a bit of calculus