Question

A block is given an initial velocity of 5.00 m/s up a frictionless 20.0° incline. How far up the incline does the block slide before coming to rest?

Answer 1

The law of conservation of energy says

$$\frac{mv^2}{2}=mgh=mg l\sin\theta$$

Hence

$$l=\frac{v^2}{2g\sin\theta}=\frac{5.00^2}{2*9.8\sin20.0^\circ}=3.73\:\rm m$$