Question #8f009

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Question #8f009

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Answer 1 · 2015-02-09T02:04:05

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Answer 2 · 2015-02-09T02:28:54

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We need to find R which is given by R= $U_{x}$ $U_x$ T

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

Using constant acceleration linear equations;

$g = \frac{V_{y} - U_{y}}{T}$ $g=(V_y-U_y)/T$ and $h_{y} = (\frac{U_{y} + V_{y}}{2}) T$ $h_y=((U_y+V_y)/2)T$
Substitute 1st equation into the second one to eliminate $V_{y}$ $V_y$ ;
$h_{y} = (2 U_{y} + g T) \frac{T}{2}$ $h_y=(2U_y+gT)T/2$ and rearrange it to get a quadratic equation,
$g T^{2} + 2 U_{y} T - 2 h_{y} = 0$ $gT^2+2U_yT-2h_y=0$

$g = 9.81 m s^{- 1}, 2 U_{y} = 2 \times 15 sin (20) m s^{- 1}, 2 h_{y} = 2 \times 28 m$ $g=9.81ms^-1, 2U_y=2xx15sin(20)ms^-1, 2h_y=2xx28m$

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

Now $R = U_{x} T = 15 cos (20) \times 1.923 m$ $R=U_xT=15cos(20)xx1.923m$
$R = 27.1 m$ $R=27.1m$

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Question #8f009

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