Answer

Verified

390.3k+ views

**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 $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$.

Recently Updated Pages

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

10 examples of friction in our daily life

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

10 examples of law on inertia in our daily life

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE