Question 964089
{{{y=(-5/72)x^2+(5/3)x}}}
By letting y=0, we can find the horizontal distance at the start (0 in traveled) and the end of the jump: 
{{{0=(-5/72)x^2+(5/3)x}}}
{{{0=x^2-24x}}}
{{{0=x-24}}}
{{{24=x}}}
The frog traveled 24 inches to the shore.
The maximum height would be mid way, when x=12 in, so
ANSWER 1: Maximum height at 12 inches horizontal distance.
{{{y=(-5/72)(12in)^2+(5/3)(12in)}}}
{{{y=(-5/72)(144)+(60/3)}}}
{{{y=(-10)+(20)}}}
{{{y=10in}}}
ANSWER 2: Maximum height was 10 inches.
This graph is sort of a picture of the jump:
{{{ graph( 800, 400, 0, 24, 0, 11, (-5/72)x^2+(5/3)x)}}}