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Question #45954

1 Answer

Mar 5, 2017

$θ \approx 42^{\circ}$ $theta ~~ 42^@$

Explanation:

Note that it takes $\frac{30}{2} = 15$ $30/2 = 15$ seconds for the vertical component of the velocity to become $0$ $0$ .

Denoting the initial vertical component of the velocity by $v$ $v$ and using $g = 9.8 m s^{- 2}$ $g = 9.8 ms^(-2)$ , we have:

$v = 9.8 \cdot 15 = 147$ $v = 9.8 * 15 = 147$

So the launch angle $θ$ $theta$ satisfies:

$sin (θ) = \frac{147}{220}$ $sin(theta) = 147/220$

So $θ = {sin}^{- 1} (\frac{147}{220}) \approx 42^{\circ}$ $theta = sin^(-1)(147/220) ~~ 42^@$

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