The speed of sound at T=25°C is Vs=346 m/s. So the sound has to reach the cliff and return back to you so the path it needs to travel is s=2*440 m = 880 m.
Since the speed of sound is constant s=Vs*t, and t= s/Vs=880/346=2.54335 s. You will hear the echo after t=2.54335 s after you shouted.
According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.
To solve this problem we must apply the concept related to the conservation of energy theorem.
PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so



PART B) Replacing the values given as,




Therefore the speed of the masses would be 1.8486m/s
I think it’s A protons .. hope this helps
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.