Ronald L. answered • 09/30/23
Over 20 years High School Physics Teacher Extraordinaire M.Ed., Ed.S.
I agree with Rachel L's solution. I would like to add an explanation.
This is a 2D problem, and we should assume that air is neglected. The 2 dimensions of horizontal and vertical aspects are completely independent of each other. In projectile motion, you have two dimensions that we can discuss. The horizontal dimension is non-accelerated and moves at constant speed. Therefore, the main equation that you would use is vx=Δx/Δt ; however, the equations that would be used for the vertical dimension would be your acceleration kinematics equations, and, since the cliff diver problem is considered the "special case" of a projectile launch (i.e., an object launched horizontally), you can assume that vertical initial velocity is zero therefore Δy=.5*g*Δt2 or Δt=(2Δy/g)0.5
Therefore, to solve the problem, 1st: figure out how long it takes, Δt, for the projectile to fall 37.1 m using the latter time equation given Δy = –37.1 m Next, calculate initial velocity using the equation vx=Δx/Δt where Δx would be set at 11.39 meters. By doing so, you would be calculating the horizontal distance that you need to avoid hitting the rock.
If we look into this problem a bit further (pun not intended), there is an important assumption that you need to make. That is, the height of the person is part of the height of the cliff. If you carried this out in real life, I am thinking that the person's leg (if falling straight) would hit the water; however, the rest of the body may turn and hit the rock! If I were given a cliff diver advice on doing this in real life, subtract the person's height from the range first, Δx. However, most of these types of problems would consider a person's height negligible for the purpose of solving...
