A) \[19.6\text{ }m{{s}^{-1}},\]vertical
B) \[24.5\text{ }m{{s}^{-1}},\]vertical
C) \[19.6\text{ }m{{s}^{-1}},53{}^\circ \]with the road
D) \[24.5\text{ }m{{s}^{-1}},53{}^\circ \]with the road
Correct Answer: D
Solution :
\[{{U}_{x}}=14.7\text{ }m/s;\text{ }{{U}_{y}}=9.8\times 2=19.6\text{ }m/s\] \[u\text{ }=\sqrt{{{\left( 14.7 \right)}^{2}}+{{\left( 19.6 \right)}^{2}}}=24.5\text{ }m/s\]\[\tan \,\theta =\frac{19.6}{14.7}=\frac{4}{3}\Rightarrow \theta =53{}^\circ \]
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