Asled starts from rest,
and slides down a 38.0°
frictionless hill for 4.24 s.
how fast is it going at
the end?

25.6 m/s

Explanation:

Draw a free body diagram of the sled.  There are two forces acting on the sled:

Normal force pushing perpendicular to the hill

Weight force pulling straight down

Take sum of the forces parallel to the hill:

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 38.0°)

a = 6.03 m/s²

Given:

v₀ = 0 m/s

a = 6.03 m/s²

t = 4.24 s

Find: v

v = at + v₀

v = (6.03 m/s²) (4.24 s) + (0 m/s)

v = 25.6 m/s