rrect Option is (A)

To find the velocity of the projectile at t = 2 s, we need to find the horizontal and vertical components of the velocity at that time.

The initial horizontal component of the velocity is constant and is given by:

$v_{0x}=v_0\cos \theta =40{\mathrm{ms}}^{-1}\cos ({30}^{\circ })=40{\mathrm{ms}}^{-1}\cdot \frac{\sqrt{3}}{2}=20\sqrt{3}{\mathrm{ms}}^{-1}$

The initial vertical component of the velocity is:

$v_{0y}=v_0\sin \theta =40{\mathrm{ms}}^{-1}\sin ({30}^{\circ })=40{\mathrm{ms}}^{-1}\cdot \frac{1}{2}=20{\mathrm{ms}}^{-1}$

To find the vertical component of the velocity at t = 2 s, we use the equation:

$v_y=v_{0y}-gt=20{\mathrm{ms}}^{-1}-(10{\mathrm{ms}}^{-2})(2\mathrm{s})=20{\mathrm{ms}}^{-1}-20{\mathrm{ms}}^{-1}=0{\mathrm{ms}}^{-1}$

At t = 2 s, the horizontal component of the velocity is still $20\sqrt{3}{\mathrm{ms}}^{-1}$, and the vertical component is 0. The overall velocity at t = 2 s is:

$\overrightarrow{v}=20\sqrt{3}{\mathrm{ms}}^{-1}\hat{i}+0{\mathrm{ms}}^{-1}\hat{j}=20\sqrt{3}{\mathrm{ms}}^{-1}$

So the correct answer is: $20\sqrt{3}{\mathrm{ms}}^{-1}$

rrect Option is (C)

Let the initial speed of both projectiles be v. The maximum height attained by a projectile can be calculated using the formula:

$H=\frac{v^2{\sin }^2\theta }{2g}$

where H is the maximum height, v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

For the projectile projected at 30°, the maximum height is:

$H_1=\frac{v^2{\sin }^2{30}^{\circ }}{2g}=\frac{v^2\times \frac{1}{4}}{2g}=\frac{v^2}{8g}$

For the projectile projected at 60°, the maximum height is:

$H_2=\frac{v^2{\sin }^2{60}^{\circ }}{2g}=\frac{v^2\times \frac{3}{4}}{2g}=\frac{3v^2}{8g}$

Now, let's find the ratio of the maximum heights:

$\frac{H_1}{H_2}=\frac{\frac{v^2}{8g}}{\frac{3v^2}{8g}}=\frac{v^2}{3v^2}$

The v² terms cancel out, and we get:

$\frac{H_1}{H_2}=\frac{1}{3}$

Therefore, the ratio of the maximum heights attained by the two projectiles is 1 : 3

rrect Option is (B)

The range $R$ of a projectile launched with an initial speed $v$ and at an angle $\theta$ to the horizontal is given by:

$R=\frac{v^2}{g}\sin (2\theta )$

where $g$ is the acceleration due to gravity.

From this equation, we can see that the range is dependent on the sine of twice the launch angle.

Given that the range at ${15}^{\circ }$ is $50$ m, if we launch the projectile at ${45}^{\circ }$ with the same velocity, we can compare the ranges by comparing $\sin (2\times {15}^{\circ })$ and $\sin (2\times {45}^{\circ })$:

$\sin ({30}^{\circ })=\frac{1}{2}$

$\sin ({90}^{\circ })=1$

Therefore, the range at ${45}^{\circ }$ will be twice the range at ${15}^{\circ }$, because $\sin ({90}^{\circ })$ is twice as large as $\sin ({30}^{\circ })$.

So, the range when the projectile is launched at ${45}^{\circ }$ will be $2\times 50$ m = $100$ m.

So, $100$ m is the correct answer.

rrect Option is (B)

The equation of the trajectory given is $y=x-\frac{x^2}{20}$.

This is a parabola, and it represents the path of the projectile.

The maximum height of the projectile corresponds to the vertex of the parabola.

The x-coordinate of the vertex for a parabola given by $y=a{x}^2+bx+c$ is $-b/2a$.

In this case, $a=-1/20$ and $b=1$, so the x-coordinate of the vertex is:

$x_{\text{vertex}}=-\frac{b}{2a}=-\frac{1}{2\times (-1/20)}=10$

Substituting this into the equation of the trajectory gives the y-coordinate of the vertex, which is the maximum height:

$y_{\text{max}}=10-\frac{{10}^2}{20}=10-5=5$

So, the maximum height attained by the projectile is 5 m.

rrect Option is (A)

The range of a projectile launched with an initial velocity $v$ at an angle $\theta$ with respect to the horizontal is given by:

$R=\frac{v^2\sin (2\theta )}{g}$,

where $g$ is the acceleration due to gravity.

Let's calculate the ranges of projectiles A and B:

For projectile A, $v=40\text{m/s}$ and $\theta ={30}^{\circ }$, so:

$R_A=\frac{(40{)}^2\sin (2\times 30)}{10}=4\times 40=160\text{m}$.

For projectile B, $v=60\text{m/s}$ and $\theta ={60}^{\circ }$, so:

$R_B=\frac{(60{)}^2\sin (2\times 60)}{10}=6\times 60=360\text{m}$.

Therefore, the ratio of their ranges is $R_A:R_B=160:360=4:9$.