Physics › Motion In Two Dimensions › Projectile motion
The equation of trajectory of a projectile is $y=10 x-\left(\frac{5}{9}\right) x^2$. If we assume $g=10…
The equation of trajectory of a projectile is $y=10 x-\left(\frac{5}{9}\right) x^2$.
If we assume $g=10 \mathrm{~ms}^{-2}$, the range of projectile (in metre) is
36 24 18 9 Solution
Equation of projectile is
$y=10 x-\left(\frac{5}{9}\right) x^2$
Standard equation is
$y=x \tan \theta-\frac{g}{2 u^2 \cos ^2 \theta} \cdot x^2$
On comparing,
$\tan \theta=10$
and
$\begin{array}{rlrl}
& \frac{g}{2 u^2 \cos ^2 \theta} & =\frac{9}{9} \\
\Rightarrow & 10 u^2 \cos ^2 \theta & =9 \mathrm{~g} \\
& \ddots & g & =10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore & u^2 \cos ^2 \theta & =9
\end{array}$
(given)
$\therefore$ Range of projectile $R=\frac{2 u^2 \sin \theta \cos \theta}{g}$
$\begin{aligned}
& =\frac{2 u^2 \tan \theta \cdot \cos ^2 \theta}{g} \\
& =\frac{2\left(u^2 \cos ^2 \theta\right) \cdot \tan \theta}{g} \\
& =\frac{2 \times 9 \times 10}{10}=18 \mathrm{~m}
\end{aligned}$
Asked in: AP EAMCET 2005
Practice more Motion In Two Dimensions questions on Aicharya