# Projectile with friction on an inclined plane

by kraaaaamos
Tags: friction, inclined, plane, projectile
 P: 218 It is simpler to resolve velocities and forces parallel and perpendicular to the ramp rather than horizontal and vertical. Reason: the block moves parallel to the ramp; the normal force (required to calculate the frictional force) is perpendicular to the ramp. Using this approach I get this expression for the answer to a $$\frac{10^{2}tan35}{2g(sin35 + 0.2cos35}$$