Question 936613
Assume that (0,0) is the position of the thrower's feet.
The ball is released at {{{x=0}}} and {{{y=5}}}.
x along the ground, y positive up, gravity acts downwards.
{{{x(t)=V[0]cos(58)t}}}
{{{x(t)=48.75t}}}
.
.
.
{{{y(t)=5+V[0]sin(58)t-(1/2)(32.2)t^2}}}
{{{y(t)=5+78t-16.1t^2}}}