# Question #8f009

Feb 9, 2015

Feb 9, 2015

We need to find R which is given by R=${U}_{x}$T

However, we need to find time of flight, T.

Using constant acceleration linear equations;

$g = \frac{{V}_{y} - {U}_{y}}{T}$ and ${h}_{y} = \left(\frac{{U}_{y} + {V}_{y}}{2}\right) T$
Substitute 1st equation into the second one to eliminate ${V}_{y}$;
${h}_{y} = \left(2 {U}_{y} + g T\right) \frac{T}{2}$ and rearrange it to get a quadratic equation,
$g {T}^{2} + 2 {U}_{y} T - 2 {h}_{y} = 0$

$g = 9.81 m {s}^{-} 1 , 2 {U}_{y} = 2 \times 15 \sin \left(20\right) m {s}^{-} 1 , 2 {h}_{y} = 2 \times 28 m$

You will get $T = 1.923 s$ (take the positive one).

Now $R = {U}_{x} T = 15 \cos \left(20\right) \times 1.923 m$
$R = 27.1 m$