The velocity of water can be decomposed into its vertical and horizontal components:
The vertical component will exhibit a parabolic trajectory due to gravity, while the horizontal component will be linear:
To determine when the water reaches the ground 2.5m away, set y= 0 and x = 2.5
According to Newton's second law, Force equals the rate of change of momentum over time. Momentum change is equal to Force times time. So, F=ma can be rearranged to a=F/m, a more recognizable formulation of Newton's second law
Using a relevant kinematic equation for mass m: V=u+at; where initial speed u=0; thus, acceleration a=F/m gives V=(F/m)xt, which translates to t=mV/F. For mass 2m, applying the same formula: V=u+at; u=0; a=F/2m indicates V=(F/2m)xt, leading to t=2mV/F (possibly double the initial time)
I might have erred somewhere along the line, but the fundamental concept seems valid... using another kinematic equation for m: s=ut + (1/2)at²; with s=d; and initial speed u=0; a=F/m; t=1; results in d=(1/2)(F/m) = F/2m. Similarly, for 2m: s=ut + (1/2)at²; s=d; u=0; a=F/2m; and t=1 gives d=(1/2)(F/2m)=F/4m (half the distance perhaps???? WHAT???!)
The amount of work performed by a system at consistent pressure is defined by the following equation:
where
p represents pressure
as the final volume
as the initial volume
Plugging the values given in this case into the formula gives us
Considering that
, the result for the work done becomes
