Question 824808
IF THIS IS PHYSICS:
Since movement is along a straight line, we can indicate direction (upwards or downwards) by assigning a positive sign to measures of displacement, velocity, and acceleration in one direction. Displacement, velocity, and acceleration in the opposite direction are assigned a negative sign. It does not matter which direction is taken as the positive direction, as long as you do not forget halfway through the problem.
 
I say upwards is the positive direction.
I say position 0 is the start of the leap,
and position 1 is at the top of the leap.
{{{S=1.07m}}} is the positive displacement of the leaper (distance from floor to feet) at the highest point of the leap, position 1.
{{{t}}} is the time rising to the top of the leap,
{{{g=-9.8}}}{{{"m/" s^2}}} is the acceleration (negative because it is downwards),
{{{v[0]>0}}} is the velocity at the start of the leap, position 0.
{{{v[1]=0}}} is the velocity at the top of the leap, position 1.
 
IF THIS IS JUST MATH:
You've been given two magical formulas:
{{{v[1]^2=v[0]^2+2gS}}} and {{{S=v[0]t+(1/2)gt^2}}} .
Since you do not know {{{t}}} , you'll have to use {{{v[1]^2=v[0]^2+2gS}}}
Substituting the values listed above, with meters and seconds in the units,
{{{0^2=v[0]^2+2(-9.8)*1.07}}}
{{{0=v[0]^2-20.972}}}
{{{v[0]^2=20.972}}}
{{{v[0]=sqrt(20.972)}}}
{{{highlight(v[0]=4.6)}}}{{{"m/s"}}}
 
IF THIS IS PHYSICS:
{{{v[1]=v[0]+gt}}} and {{{S=v[0]t+(1/2)gt^2}}}
do not seem useful by themselves
{{{(v[1]+v[0])/2}}} is the average velocity, so {{{S=(v[1]+v[0])t/2}}} .
From there we can get to {{{S=v[0]t+(1/2)gt^2}}}, but we can also get to {{{v[1]^2=v[0]^2+2gS}}} .
{{{S=(v[1]+v[0])t/2}}}-->{{{2S=(v[1]+v[0])t}}}-->{{{t=2S/(v[1]+v[0])}}}
Substituting {{{2S/(v[1]+v[0])}}} for {{{t}}} into {{{v[1]=v[0]+gt}}} we get
{{{v[1]=v[0]+2gS/(v[1]+v[0])}}}
{{{v[1]-v[0]=2gS/(v[1]+v[0])}}}
{{{(v[1]-v[0])(v[1]+v[0])=2gS}}}
{{{v[1]^2-v[0]^2=2gS}}}
{{{v[1]^2=v[0]^2+2gS}}}