Answer
Yes, the ball is a home run.
Work Step by Step
$x=x_0+(v_0 \cos \theta )t=400$( $x$-value of the ball)
or, $0(115 \cos 50^\circ )t=400$
Thus, $t=\frac{400}{(115 \cos 50^\circ )}=5.41$ s
Now, $y=y_0+(v_0 \sin \theta )t-\dfrac{1}{2}gt^2$ ($y$-value of the ball.)
Thus,$y=3+(115 \sin 50^\circ )(5.41)-\dfrac{1}{2}(9.8)(5.41)^2$
or, $y= 11.3 ft$
Now, $t=5.41 s$($y$-position of the ball is 11.3 ft.)
We can see that $11.3 ft \gt 10 ft$
From the above calculations, we conclude that the ball is a home run.
