A 1.80-kg block slides down a frictionless ramp.The top of the ramp is h1 = 1.21 m above the ground; the bottom of the ramp is h2 = 0.256 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance $d$ away. Calculate the distance d.

Answer 1

(1/2) m u^2 = m g (Hup - Hdown)
so
u = sqrt(2*9.81*.954)
u is horizontal speed, does not change

How long in the air? Well, how long to fall .256 m ?
.256 = 4.9 t^2
solve for t, time in the air
distance horizontal = d = u t