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The trajectory of a projectile projected from origin is given by the equation \( y=x-\frac{2 x^{2}}{5} \). The
initial velocity of the projectile is
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initial velocity of the projectile is
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Verified Answer
The correct answer is:
\( 5 \mathrm{~ms}^{-1} \)
\( (D) \)
\[
\begin{array}{l}
y=x-\frac{2 x^{2}}{5} \\
\tan \theta=1 \\
\theta=45^{\circ} \\
\frac{g}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5} \\
\frac{10}{2 u^{2} \frac{1}{2}}=\frac{2}{5} \\
u^{2}=\frac{50}{2}=25 \\
u=5 m s^{-1}
\end{array}
\]
\[
\begin{array}{l}
y=x-\frac{2 x^{2}}{5} \\
\tan \theta=1 \\
\theta=45^{\circ} \\
\frac{g}{2 u^{2} \cos ^{2} \theta}=\frac{2}{5} \\
\frac{10}{2 u^{2} \frac{1}{2}}=\frac{2}{5} \\
u^{2}=\frac{50}{2}=25 \\
u=5 m s^{-1}
\end{array}
\]
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