Answer
92 m
Work Step by Step
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the flight time in the first incident.
$\uparrow S=ut+\frac{1}{2}at^{2}$
$0=Vsin\theta\times t+\frac{1}{2}(-g)t^{2}$
$t=\frac{2Vsin\theta}{g}-(1)$
Let's apply equation $S=ut+\frac{1}{2}at^{2}$ in the horizontal direction to find the range in the first incident.
$R=Vcos\theta\times t$
(1)=>
$R=Vcos\theta\times (\frac{2Vsin\theta}{g})$
$R=\frac{V^{2}sin2\theta}{g}-(2)$
$23\space m=\frac{V^{2}sin2\theta}{g}-(3)$
When launch speed doubles V=>2V
(2)=>
$R_{new}=\frac{(2V)^{2}sin2\theta}{g}$
$R_{new}=\frac{4V^{2}sin2\theta}{g}$
(3)=>
$R_{new}=4\times 23\space m=92\space m$
So, the new range is 92 m
