Answer:
The baseball will be 42 meters horizontally from where it was thrown and at the heigth of the ground.
Explanation:
this is a problem that involves basic mechanics in two directions, for the x axis we have a constant speed problem and for the y axis we have a constant aceleration problem, using the equations for each one of these we got:
[tex]X=Xo+Vox*t[/tex]
[tex]Y=Yo+Voy*t+\frac{a*t^{2} }{2}[/tex]
where Xo and Yo are the inital position for x and y axis, in this case [tex]Xo=0[/tex] and [tex]Yo =9.6m[/tex], Vox and Voy are the initial speed for x and y axis, in this case [tex]Vox=30m/s[/tex] and [tex]Voy=0m/s[/tex] since the ball was throwns horizontally, and a is the acceleration in the y axis which is the gravitational aceleration, that means [tex]a=-9.8m/s^{2}[/tex](the minus sign is because the gravitational aceleration is towards the negative y axis).
Replacing these values we have to equations that will help us find the position of the ball at any time we want:
[tex]X=0+30m/s*t[/tex]
[tex]Y=9.6m+0*t-\frac{9.8m/s^{2}*t^{2}}{2}=9.6m-\frac{9.8m/s^{2}*t^{2} }{2}[/tex]
and here we just have to replace the time value and solve, so:
[tex]X=0+30m/s*1.4s=42m[/tex]
[tex]Y=9.6m-\frac{9.8m/s^{2}*1.4s^{2} }{2}=9.6m-4.9m/s^{2}*1.96s^{2}=9.6m-9.6m=0m[/tex]
and that's it, the ball is 42 meters horizontally from where it was throwns and at the heigth of the ground.
