We have the distance the hammer goes along with the angle it leaves the hand with, so we make a drawing of the projectile motion: change in x is 81.12m, change in y is -1.00m
We also make a drawing of the breakdown of the initial velocity into its x and y components:
We know that in the x direction there is no acceleration because we are neglecting air resistance. We start with the x direction and find an equation with what we know.
Now we move on to the y direction. The acceleration will be -g and we have to keep in mind the 1 meter above ground that the hammer is released at. In the third equation I have replaced the y component of velocity with the relation found from the breakdown of the initial velocity containing the velocity in the x direction. Below that, I have replaced the velocity in the x direction with the expression found directly above.
Now that we know the time that the hammer was in the air, we can find the x component of the initial velocity:
Now we can find the initial velocity magnitude using trig on the triangle at the top of the solution.
Now that we know the initial velocity, we can find the rotational velocity that corresponds to this translational velocity:
Now we just need to convert this number to revolutions per second.
