Question 971350
{{{h=3}}}= initial height of the ball above the ground (in feet)
{{{V=120}}}= initial ball speed (in feet per second)
{{{g=32}}}{{{ft}}}{{{" /"}}}{{{second^2}}}= acceleration of gravity
{{{theta=30^o}}}= initial angle of the ball trajectory with the horizontal ground
{{{t}}}= time the ball is in the air (in seconds)
{{{X=(V*cos(theta))*t}}}= horizontal displacement of the ball (in feet)
{{{Y=h+(V*sin(theta))*t-16t^2}}}= height of the ball above the horizontal ground (in feet)
 
When the ball hits the horizontal ground, {{{Y=0}}} . Then,
{{{h+(V*sin(theta))*t-16t^2=0}}} , and substituting the known values,
{{{3+(120*sin(30^o))*t-16t^2=0}}}--->{{{3+(120*(1/2))*t-16t^2=0}}}--->{{{3+60t-16t^2=0}}}--->{{{16t^2-60t-3=0}}}
The solution to that quadratic equation that makes sense for this situation is
{{{t=aproximately3.8}}} ,
because the other solution is negative
Plugging that value for {{{t}}} ,
and the known values for {{{theta}}} and {{{V}}} into
{{{X=(V*cos(theta))*t}}} we get
{{{X=(120*cos(30^o))*3.8}}}--->{{{X=approximately395}}}