22P1Q02

2 C

Vertically: (time of flight)

$\displaystyle \begin{aligned} {{t}_{f}}&=\frac{2{{u}_{y}}}{g} \\ 2.0&=\frac{2{{u}_{y}}}{9.81} \\ {{u}_{y}}&=9.81\text{ m }{{\text{s}}^{-1}} \end{aligned}$

Horizontally:

$\displaystyle \begin{aligned} {{s}_{x}}&={{u}_{x}}{{t}_{f}} \\ 100&={{u}_{x}}(2.0) \\ {{u}_{x}}&=50\text{ m }{{\text{s}}^{-1}} \end{aligned}$

$\displaystyle \begin{aligned} \tan \theta &=\frac{{{u}_{y}}}{{{u}_{x}}}\\&=\frac{9.81}{50} \\ \theta &=11.1{}^\circ \end{aligned}$

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