Answer
The ball would land $~~3.34~m~~$ farther away.
Work Step by Step
We can find the time of flight if the angle is $\theta = 18.00^{\circ}$:
$y = v_{y0}~t+\frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_{y0}~t-y = 0$
$4.9~t^2+(20.0~sin~18.00^{\circ})~t-2.30 = 0$
$4.9~t^2+6.18~t-2.30 = 0$
We can use the quadratic formula:
$t = \frac{-6.18\pm \sqrt{(6.18)^2-(4)(4.9)(-2.30)}}{(2)(4.9)}$
$t = \frac{-6.18\pm \sqrt{83.2724}}{9.8}$
$t = -1.56~s, 0.301~s$
Since $t$ is positive, we can choose the positive solution.
We can find the horizontal distance the ball travels:
$x = (20.0~m/s)~(cos~18.00^{\circ})(0.301~s)$
$x = 5.73~m$
We can find the time of flight if the angle is $\theta = 8.00^{\circ}$:
$y = v_{y0}~t+\frac{1}{2}gt^2$
$\frac{1}{2}gt^2+v_{y0}~t-y = 0$
$4.9~t^2+(20.0~sin~8.00^{\circ})~t-2.30 = 0$
$4.9~t^2+2.78~t-2.30 = 0$
We can use the quadratic formula:
$t = \frac{-2.78\pm \sqrt{(2.78)^2-(4)(4.9)(-2.30)}}{(2)(4.9)}$
$t = \frac{-2.78\pm \sqrt{52.8084}}{9.8}$
$t = -1.03~s, 0.458~s$
Since $t$ is positive, we can choose the positive solution.
We can find the horizontal distance the ball travels:
$x = (20.0~m/s)~(cos~8.00^{\circ})(0.458~s)$
$x = 9.07~m$
We can find the difference in horizontal distance:
$\Delta x = (9.07~m)-(5.73~m) = 3.34~m$
The ball would land $~~3.34~m~~$ farther away.
