Answer
1.7 m/s
Work Step by Step
Since there is no acceleration in the horizontal direction, the x-component of velocity remains constant for both balls.
Let's apply the equation $S=ut$ in horizontal directions for balls A & B.
Ball (A)=>
$x_{A}=V_{xA}t_{A}$
Ball (B)=>
$x_{B}=V_{xB}t_{B}$
Given that, $x_{A}=x_{B}$ So we get.
$V_{xA}t_{A}=V_{xB}t_{B}=> V_{xB}=V_{xA}(\frac{t_{A}}{t_{B}})-(1)$
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in vertical direction for both balls A & B.
$\uparrow S=ut+\frac{1}{2}at^{2}$
$-y_{A}=0+\frac{1}{2}(-g)t_{A}^{2}$
$t_{A}=\sqrt {\frac{2y_{A}}{g}}-(2)$
Similarly,
$t_{B}=\sqrt {\frac{2y_{B}}{g}}-(3)$
(2),(3)=>(1),
$V_{xB}=V_{xA}(\frac{\sqrt {\frac{2y_{A}}{g}}}{\sqrt {\frac{2y_{B}}{g}}})=V_{xA}\sqrt {\frac{y_{A}}{y_{B}}}$
$V_{xB}=1.9\space m/s \sqrt {\frac{1.2\space m}{1.5\space m}}=1.7\space m/s$
