Question

A 5 kg present is initially at rest on the floor: μ = 0.25. A force of 120 N is then exerted onto the present, directed at an angle of 30∘ above the horizontal. The applied force is exerted onto the present for 3.0 s, at which point the applied force is removed. The present then moves as a projectile. It lands on the floor without bouncing and then slides to a stop. How far does the present end up from where it started?

Answer 1

$$F\cos{30}t=mv\\120(3)\cos{30}=5v\\v=62.35\frac{m}{s}$$

$$v^2=2\mu g d\\62.35^2=2(0.25)(9.8)d\\d=793\ m$$