It's been a great day of new, frictionless snow. julie starts at the top of the 60∘ slope shown in the figure. at the bottom, a circular arc carries her through a 90∘ turn, and she then launches off a 3.0-m-high ramp. what horizontal distance will she travel before touching down?

She travels approximately 40m before touching down.

Equating inital and final kinetic and potential energies,

m + mg = m + mg

Solving for velocity, at the end of the ramp, v = sqrt()

Now, velocity as she is going off the ramp should be

= sqrt( = 20.8 m/s

Now,  y = + t - g    
            = 3 + 20.8 sin (60) - 4.9
            = 3 + 18.01t - 4.9 
            t = 3.83
Now, distance x = speed*time 
                            =  vt = cos(θ)t
                            = 20.8 * cos(60) * 3.83
                            = 40 m

The correct answer should be b let me know if i’m wrong