Answer
The speed of the baseball must be 91.5 mph
Work Step by Step
Let $t$ be the time it takes for the ball to reach home plate. In general:
$t = \frac{d}{v}$
Let $v_b$ be the required speed of the baseball:
$t = \frac{43.0~ft}{65.0~mph} = \frac{60.5~ft}{v_b}$
$v_b = (\frac{60.5~ft}{43.0~ft})~(65.0~mph)$
$v_b = (1.407)~(65.0~mph)$
$v_b = 91.5~mph$
The speed of the baseball must be 91.5 mph
