1. The problem statement, all variables and given/known data A projectile starts from rest and moves 4.7 m down a frictionless ramp inclined at 21 degrees with the horizontal. the acceleration due to gravity is 9.8 m/s^2 what what speed will it leave the ramp ? what will the range of the projectile if the bottom of the ramp is 2.1 m above the ground 2. Relevant equations V=Vi + at delta X = Vi*t+1/2 a*t^2 Vf^2 - Vi^2 = 2*a*deltaX 3. The attempt at a solution Vertical, Viy=0 ay=-9.8 Vfy= - Vf sin 21 delta y= -4.7 sin 21 horizontal, Vix= ( i thought it to be zero at first but it cant be zero because then the delta x comes out to be zero. So im confused at this one ) ax=0 Vfx=Vf cos 21 delta x= 4.7 cos 21 so i used the third law and set it up as follows, Vfy^2 = 2 (-9.8) (4.7 sin 21) then the idea is to set the answer = Vf sin 21 and solve for Vf it comes out to be 16.0329 but this is unfortunately not the right answer. can anyone help ?