A car of mass 800 kg is accelerated horizontally by constant net force of 1920 N for 9 s. It then breaks for 2 s, but drives off a 5 m high cliff. If μ = 0.85, what is the total horizontal distance travelled by car and its velocity? Ignore air resistance.

F=ma, so a=F/m giving a=2.4m/s^2. When t=9s, v=at, so v=21.6m/s. Friction force is Ff = μF(normal), which is Ff = μmg, so Ff = 6670.8N (if g = 9.81m/s^2). Frictional force always acts against the direction of motion. Deceleration due to frictional force d = -Ff/m = -μg = -8.34m/s^2. Speed after 2 seconds of deceleration = 4.9m/s. In free fall we can consider horizontal and vertical velocity components separately. Vertical: need time in which the car falls 5m. h = 1/2gt^2, so t = sqrt(2h/g), which is 1.04s. Horizontal: speed doesn't change, so the distance travelled is s = vt, s = 5.1m. Total horizontal distance: 1/2at^2 + vt + 1/2dt^2 + 5.1 = 128.82m. Considering velocity, we have to calculate both its magnitude and its direction, since velocity is a vector. At the end of free fall horizontal component didn't change, so it's equal to 4.9m/s, while vertical speed gained is v = gt, v = 10.2m/s. The magnitude of the final velocity is therefore sqrt((4.9)^2+(10.2)^2). Its direction is tan(θ) = opp/adj, giving us θ = 26°, or 154° away from normal.