Work is (force) times (distance). For Amy, you know both of them, and you can easily multiply them to find the amount of work. For Joe, the distance is zero, which should tell you all you need to know.
Answer:
ooh thanks but don't give me advise
Ok, this is a 2d kinematics problem, the falls 14 m part is confusing, I think it means in the x direction, but you don't need it anyway.
If we know it goes 4m into the air, we know d = 4m (height of wall), we also know the acceleration a=-9.8m/s^2 (because gravity) and that the vertical velocity when it just clears the wall will be 0 m/s, which we'll call our final velocity (Vf). Using Vf^2 = Vi^2 +2a*d, we can solve this for Vi and drop Vf because it's zero to get: Vi = sqrt(-2ad), plug in numbers (don't forget a is negative) and you get 8.85 m/s in the vertical direction. The x-direction velocity requires that we solve the y-direction for time, using Vf= Vi + at, we solve for t, getting t= -Vi/a, plug in numbers t= -8.85/-9.8 = 0.9 s. Now we can use the simple v = d/t (because x-direction has no acceleration (a=0)), and plug in the distance to the wall and the time it takes to get there v = (4/.9) = 4.444 m/s, this is the velocity in the x direction, we use Pythagoras' theorem to find the total velocity, Vtotal = sqrt(Vx^2 + Vy^2), so Vtotal = sqrt(8.85^2+4.444^2) = 9.9m/s. Yay physics!
