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A particle is projected in air from origin with speed $u$ and at an angle $\theta $ with positive $x - axis$. At its maximum height its trajectory has radius of curvature $R$ and centre of curvature has coordinate $\left( {20m,\dfrac{{70}}{3}m} \right)$ choose the correct option(s) $(g = - 10m{s^{ - 2}})$.
A. Equation of trajectory of the projectile motion is $y = 3x - \dfrac{{3{x^2}}}{{40}}$
B. The maximum height of the projectile is $30m$
C. The value of $u$ is $\sqrt {\dfrac{{2000}}{3}} m{s^{ - 1}}$
D. The value of $\theta $ is ${\tan ^{ - 1}}\left( {\dfrac{3}{{\sqrt {10} }}} \right)$

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: In order to find the solution of the given question, we need to know about the formula for the horizontal range and the maximum height in case of a projectile motion. Also, we need to know the values of the horizontal range and height at the centre of the curvature. After that we need to solve the equations obtained. Then we can finally conclude with the correct solution for the given question.

Complete step by step answer:
Step one:
We know that the range, $r = \dfrac{{{v^2}}}{g}$
 Since, the particle is covering the horizontal distance, we need to take the horizontal component of the velocity. Therefore, the range can be written as,
$r = \dfrac{{{{(u\cos \theta )}^2}}}{g} = \dfrac{{{u^2}{{\cos }^2}\theta }}{g}$ -----(i)
We know that the radius of curvature can be written as, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ ----(ii)
and the maximum height can be written as, $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ ----(iii)

Step two:
We also know that the coordinates at the centre of curvature are given by $\left( {\dfrac{R}{2},H - r} \right)$
Comparing the above coordinates with the values given in the question, we get,
$\dfrac{R}{2} = 20$
$ \Rightarrow R = 40$
Now comparing the above value with equation (ii), we get,
$40 = \dfrac{{{u^2}\sin 2\theta }}{g}$
$ \Rightarrow 40 = \dfrac{{{u^2}2\sin \theta \cos \theta }}{{10}}$ -----(iv)
Now, $\sin \theta = \dfrac{3}{{\sqrt {10} }}$ and $\cos \theta = \dfrac{1}{{\sqrt {10} }}$ also $\tan \theta = 3$
Putting these values in equation (iv), we get,
$40 = \dfrac{{{u^2}2 \times \dfrac{3}{{\sqrt {10} }} \times \dfrac{1}{{\sqrt {10} }}}}{{10}}$
$ \Rightarrow 400 = \dfrac{{6{u^2}}}{{10}}$
$ \Rightarrow {u^2} = \dfrac{{4000}}{6}$
$\therefore u = \sqrt {\dfrac{{4000}}{6}} = \sqrt {\dfrac{{2000}}{3}} m{s^{ - 1}}$

Step three:
Now, let us find the value of the height, $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
$ \Rightarrow H = \dfrac{{4000}}{6} \times \dfrac{9}{{10 \times 20}} = 30m$

Step four:
Now let us find the equation of $y$. It can be written as,
$y = x\tan \theta - \dfrac{1}{2} \times \dfrac{{g{x^2}}}{{{u^2}{{\cos }^2}\theta }}$
$\therefore y = 3x - \dfrac{{3{x^2}}}{{40}}$
After analyzing the values we can conclude that the options A,B,C are the correct choices for the given question.

Hence, the correct answers are option (A), (B) and (C).

Note: In a parabola, the y-intercept is a point where the parabola crosses the y-axis. We should also know this fact that the equation of a parabola in two variables can be written as $y = a{x^2} + bx + c$. The maximum height attained by a body when it is moving in a parabolic path can be found by $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$.
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