Assuming acceleration due to gravity of the moon is constant and there’s no initial velocity in the mans jump we can use one of the kinematic equations. x(final)=x(initial)+(1/2)gt^2. Plug in known values. 0=10-(1.62/2)t^2. The value 1.62 is acceleration of gravity on the moon. Now simply solve for t. t=3.513
The answer is 30 ... same as the Atomic number.
