A ball of 4kg is projected horizontally with a speed of 20m/s from the top
of a 45m tall building. What is the kinetic energy of the ball just after 2seconds?
at the top of the building, PE=mgh = 1765.8 J
after 2 seconds, the distance fallen is 4.9t^2 = 19.62m
so the height is now 45-19.62 = 25.38m
Now you know that
PE = mgh = 995.9 J
so KE = 1765.8-995.9 = 769.9 J
to check this, note that
v = at = 9.81*2 = 19.62
and so KE = 1/2 mv^2 = 769.9 J