Motion in a Plane

$$\begin{gathered} K{E}_{\text{initial }}=\frac{1}{2}m{v}^2 \\ K{E}_{\text{initial }}=\frac{1}{2}\times 0.1\times (20{)}^2=20J \\ K{E}_{\text{final }}=\frac{1}{2}m{v}_x^2 \\ K{E}_{\text{final }}=\frac{1}{2}\times 0.1\times (10{)}^2=5J\text{ Decrease in Kinetic Energy } \\ \Delta KE=K{E}_{\text{initial }}-K{E}_{\text{final }} \\ \Delta KE=20-5=15J \end{gathered}$$

Angles of projection
${\theta }_1=45+\alpha$
${\theta }_2=45-\alpha$
Time of flight formula
$T=\frac{2v\sin \theta }{g}$
Ratio of times of flight
$\frac{T_1}{T_2}=\frac{\sin (45+\alpha )}{\sin (45-\alpha )}$
Using trigonometric identities
$\frac{T_1}{T_2}=\frac{\frac{1}{\sqrt{2}}\cos \alpha +\frac{1}{\sqrt{2}}\sin \alpha }{\frac{1}{\sqrt{2}}\cos \alpha -\frac{1}{\sqrt{2}}\sin \alpha }$
Simplifying
$\frac{T_1}{T_2}=\frac{\cos \alpha +\sin \alpha }{\cos \alpha -\sin \alpha }$
Final ratio
$\frac{T_1}{T_2}=\frac{1+\tan \alpha }{1-\tan \alpha }$

The maximum height of a projectile is given by:

$$H = \frac{u^2 \sin^2 \theta}{2g}$$​

where:

• $u$ is the initial velocity,
• $\theta$ is the angle of projection,
• $g$ is the acceleration due to gravity.

The two angles of projection given are:

1. $\theta_1 = \alpha + \frac{\pi}{4}$​
2. $\theta_2 = \alpha - \frac{\pi}{4}$​

Using the formula for maximum height:

$$H_1 = \frac{u^2 \sin^2(\alpha + \pi/4)}{2g}$$

$$H_2 = \frac{u^2 \sin^2(\alpha - \pi/4)}{2g}$$​

The required ratio is:

$$\frac{H_1}{H_2} = \frac{\sin^2(\alpha + \pi/4)}{\sin^2(\alpha - \pi/4)}$$​

Using the sine angle addition and subtraction formulas:

$$\sin(\alpha + \pi/4) = \frac{\sin \alpha + \cos \alpha}{\sqrt{2}}$$

​​ $$\sin(\alpha - \pi/4) = \frac{\sin \alpha - \cos \alpha}{\sqrt{2}}$$​

Squaring both:

$$\sin^2(\alpha + \pi/4) = \frac{(\sin \alpha + \cos \alpha)^2}{2}$$

$$\sin^2(\alpha - \pi/4) = \frac{(\sin \alpha - \cos \alpha)^2}{2}$$​

Thus, the ratio simplifies to:

$$\frac{H_1}{H_2} = \frac{(\sin \alpha + \cos \alpha)^2}{(\sin \alpha - \cos \alpha)^2}$$​

Using the identity:

$$(\sin \alpha + \cos \alpha)^2 = 1 + \sin 2\alpha$$

$$(\sin \alpha - \cos \alpha)^2 = 1 - \sin 2\alpha$$

$$\frac{H_1}{H_2} = \frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}$$​

Comparing with the given options, the correct answer is:

$$\boxed{A} \quad \frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}$$​

$$\begin{gathered} \frac{{\left(H_{\max }\right)}_1}{{\left(H_{\max }\right)}_2}=\frac{u^2{\sin }^2(45-\alpha )}{u^2{\sin }^2(45+\alpha )} \\ =\frac{{\left(\frac{1}{\sqrt{2}}\cos \alpha -\frac{1}{\sqrt{2}}\sin \alpha \right)}^2}{{\left(\frac{1}{\sqrt{2}}\cos \alpha +\frac{1}{\sqrt{2}}\sin \alpha \right)}^2} \\ =\frac{1-\sin 2\alpha }{1+\sin 2\alpha } \end{gathered}$$

$\frac{r_1}{r_2} =\frac{3}{4}$

As centripetal force F $=\frac{m{v}^2}{r}$
In order to have constant (same in this question) centripetal force

\begin{aligned}
& F_1=F_2 \\
& \frac{m_1 v_1^2}{r_1}=\frac{m_2 v_2^2}{r_2} \\
& \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{r_1}{r_2}}=\frac{\sqrt{3}}{2}
\end{aligned}

The maximum height of a projectile is given by:

$$H = \frac{u^2 \sin^2 \theta}{2g}$$​

where:

• $u$ is the initial velocity,
• $\theta$ is the angle of projection,
• $g$ is the acceleration due to gravity.

The two angles of projection given are:

1. $\theta_1 = \alpha + \frac{\pi}{4}$​
2. $\theta_2 = \alpha - \frac{\pi}{4}$​

Using the formula for maximum height:

$$H_1 = \frac{u^2 \sin^2(\alpha + \pi/4)}{2g}$$

$$H_2 = \frac{u^2 \sin^2(\alpha - \pi/4)}{2g}$$​

The required ratio is:

$$\frac{H_1}{H_2} = \frac{\sin^2(\alpha + \pi/4)}{\sin^2(\alpha - \pi/4)}$$​

Using the sine angle addition and subtraction formulas:

$$\sin(\alpha + \pi/4) = \frac{\sin \alpha + \cos \alpha}{\sqrt{2}}$$

​​ $$\sin(\alpha - \pi/4) = \frac{\sin \alpha - \cos \alpha}{\sqrt{2}}$$​

Squaring both:

$$\sin^2(\alpha + \pi/4) = \frac{(\sin \alpha + \cos \alpha)^2}{2}$$

$$\sin^2(\alpha - \pi/4) = \frac{(\sin \alpha - \cos \alpha)^2}{2}$$​

Thus, the ratio simplifies to:

$$\frac{H_1}{H_2} = \frac{(\sin \alpha + \cos \alpha)^2}{(\sin \alpha - \cos \alpha)^2}$$​

Using the identity:

$$(\sin \alpha + \cos \alpha)^2 = 1 + \sin 2\alpha$$

$$(\sin \alpha - \cos \alpha)^2 = 1 - \sin 2\alpha$$

$$\frac{H_1}{H_2} = \frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}$$​

Comparing with the given options, the correct answer is:

$$\boxed{A} \quad \frac{1 + \sin 2\alpha}{1 - \sin 2\alpha}$$​

$\overrightarrow{r}=5t^2\hat{i}-5t\hat{j}$

Velocity is given by

$$\begin{gathered} \frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}=10t\hat{i}-5\hat{j} \\ \overrightarrow{v}=10\left(2\right)\hat{i}-5\hat{j} =20 \hat{i}-5 \hat{j} \\ r = \sqrt{{20}^2 + 5^2 }= 5\sqrt{17} m/s \end{gathered}$$

At t = 2 sec

$\tan \theta =\frac{V_x}{V_y}=4$

From-ve Y-axis

$\theta ={\tan }^{-1}4$

$\begin{array}{l}\overrightarrow{a}=\frac{6}{2}\hat{k}=3\hat{k}, t=\frac{5}{3}s\\ \overrightarrow{u}=3\hat{i}+4\hat{j}\\ \overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t\\ \overrightarrow{v}=3\hat{i}+4\hat{j}+5\hat{k}\end{array}$

\begin{equation}
\begin{aligned}
& \text { Maximum height } H=\frac{v^2 \sin ^2 \theta}{2 g} \\
& 4 R=\frac{4 \pi^2 R^2}{T^2 2 g} \sin ^2 \theta \quad\left[\therefore \frac{2 \pi R}{T}=V\right] \\
& \Rightarrow \sin \theta=\sqrt{\frac{2 g T^2}{\pi^2 R}} \\
& \Rightarrow \theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{\frac{1}{2}}
\end{aligned}
\end{equation}

. The velocity at maximum height for a projectile is equal to its horizontal velocity, which remains constant throughout the motion. Horizontal velocity is given by $v_x = u \cos\theta$

For $\theta = 30^\circ$: $v_{x1} = u \cos 30^\circ = u \frac{\sqrt{3}}{2}$​​.
For $\theta = 60^\circ$:$v_{x2} = u \cos 60^\circ = u \frac{1}{2}$.

The ratio of velocities at maximum height is:

$$\frac{v_{x1}}{v_{x2}} = \frac{u \cdot \frac{\sqrt{3}}{2}}{u \cdot \frac{1}{2}} = \sqrt{3}:1.$$

Thus, the correct answer is (c) $\sqrt{3}:1$

\begin{equation}
\begin{aligned}
& \text { Maximum height } H=\frac{v^2 \sin ^2 \theta}{2 g} \\
& 4 R=\frac{4 \pi^2 R^2}{T^2 2 g} \sin ^2 \theta \quad\left[\therefore \frac{2 \pi R}{T}=V\right] \\
& \Rightarrow \sin \theta=\sqrt{\frac{2 g T^2}{\pi^2 R}} \\
& \Rightarrow \theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{\frac{1}{2}}
\end{aligned}
\end{equation}

. The velocity at maximum height for a projectile is equal to its horizontal velocity, which remains constant throughout the motion. Horizontal velocity is given by $v_x = u \cos\theta$

For $\theta = 30^\circ$: $v_{x1} = u \cos 30^\circ = u \frac{\sqrt{3}}{2}$​​.
For $\theta = 60^\circ$:$v_{x2} = u \cos 60^\circ = u \frac{1}{2}$.

The ratio of velocities at maximum height is:

$$\frac{v_{x1}}{v_{x2}} = \frac{u \cdot \frac{\sqrt{3}}{2}}{u \cdot \frac{1}{2}} = \sqrt{3}:1.$$

Thus, the correct answer is (c) $\sqrt{3}:1$

Velocity:

$\overrightarrow{v}=-x\hat{i}+2y\hat{j}-z\hat{k}$

Acceleration: $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=\left(-\frac{dx}{dt}\right)\hat{i}+2\left(\frac{dy}{dt}\right)\hat{j}-\left(\frac{dz}{dt}\right)\hat{k}$

Since $\frac{dx}{dt}=v_x=-x;\frac{dy}{dt}=2y;\frac{dz}{dt}=-z$

$\overrightarrow{a}=(-(-x\left)\right)\hat{i}+2\left(2y\right)\hat{j}-(-z)\hat{k}$

At (1, 2, 4):

$\overrightarrow{a}=1\hat{i}+8\hat{j}+4\hat{k}$

Magnitude:

$\left|\overrightarrow{a}\right|=\sqrt{1^2+8^2+4^2}=\sqrt{81}=9$

$x=24t\Rightarrow v_x=\frac{dx}{dt}=24 \mathrm{m}/\mathrm{s}$

$y=43.6t-4.9t^2\Rightarrow v_y=\frac{dy}{dt}=43.6-9.8t$

At t = 2s:

$v_y=43.6-9.8\left(2\right)=43.6-19.6=24 \mathrm{m}/\mathrm{s}$

$\tan \theta =\frac{v_y}{v_x}=\frac{24}{24}=1$

$\therefore \theta =45°$

Here horizontal range and maximum height is same. so,

\begin{equation}
\begin{aligned}
& \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{~g}}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\
& 4 \sin \theta \cos \theta=\sin ^2 \theta \\
& 4=\tan \theta
\end{aligned}
\end{equation}

$\overrightarrow{r}=5t^2\hat{i}-5t\hat{j}$

Velocity is given by

$$\begin{gathered} \frac{d\overrightarrow{r}}{dt}=\overrightarrow{v}=10t\hat{i}-5\hat{j} \\ At t = 2 sec \\ \overrightarrow{v}=10\left(2\right)\hat{i}-5\hat{j} =20 \hat{i}-5 \hat{j} \\ r = \sqrt{{20}^2 + 5^2 }= 5\sqrt{17} m/s \end{gathered}$$

$\tan \theta =\frac{v_x}{v_y}=4$

From-ve Y-axis

$\theta ={\tan }^{-1}4$

1. Initial Vertical Velocity ($u_y$):

   $$u_y = u \sin \theta = 60 \sin 30^\circ = 30 \, \text{m/s}.$$

2. Time of Flight (Total time of upward motion):
   Using the formula for the total time of motion upward:

   $$T_{\text{up}} = \frac{u_y}{g} = \frac{30}{10} = 3 \, \text{seconds}.$$

   The last second in the upward motion corresponds to the time $t = 2 \, \text{seconds} \to 3 \, \text{seconds}$

Using the equation of motion:

$$h = u_y t - \frac{1}{2} g t^2,$$

for $t = 1$:

$$h_1 = (30)(1) - \frac{1}{2}(10)(1)^2 = 30 - 5 = 25 \, \text{m}.$$

For the last second in the upward motion, the time interval is $t = 2 \to 3 \, \text{seconds}$. To find $h_2$, calculate the difference in height between these two time intervals:

$$h_2 = \text{Height at } t = 3 \, \text{seconds} - \text{Height at } t = 2 \, \text{seconds}.$$

$$h(t=2) = (30)(2) - \frac{1}{2}(10)(2)^2 = 60 - 20 = 40 \, \text{m}.$$

$$h(t=3) = (30)(3) - \frac{1}{2}(10)(3)^2 = 90 - 45 = 45 \, \text{m}.$$

$$h_2 = h(t=3) - h(t=2) = 45 - 40 = 5 \, \text{m}.$$

$$\frac{h_1}{h_2} = \frac{25}{5} = 5:1.$$

Final Answer:

$$\boxed{5:1}$$​

\begin{equation}
\begin{aligned}
& \frac{u^2 \sin 2 \theta}{g}=\frac{3 u^2 \sin ^2 \theta}{2 g} \\
& 2 \sin \theta \cos \theta=\frac{3}{2} \sin \theta \sin \theta \\
& \frac{4}{3}=\tan \theta \\
& \theta=53^{\circ} \\
& R=\frac{u^2}{g} \times 2 \times \frac{4}{5} \times \frac{3}{5}=\frac{24}{25} \frac{u^2}{g}
\end{aligned}
\end{equation}

Here horizontal range and maximum height is same. so,

\begin{equation}
\begin{aligned}
& \frac{\mathrm{u}^2 \sin 2 \theta}{\mathrm{~g}}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\
& 4 \sin \theta \cos \theta=\sin ^2 \theta \\
& 4=\tan \theta
\end{aligned}
\end{equation}

The maximum height of the projectile is given by

$$\begin{gathered} \mathrm{H} = \frac{{\mathrm{u}}^2 {\sin }^2\mathrm{\theta }}{2\mathrm{g}} \\ \mathrm{So}, \\ \frac{{\mathrm{H}}_1}{{\mathrm{H}}_2} = \frac{{\sin }^2{\mathrm{\theta }}_1}{{\sin }^2{\mathrm{\theta }}_2} = \frac{{\sin }^{2}60°}{{\sin }^{2}30°} \\ \frac{{\mathrm{H}}_1}{{\mathrm{H}}_2} =\sqrt{3} : 1 \end{gathered}$$

If two projectiles with the same speed have the same range, their angles are complementary.
Time of flight:

$T_1=\frac{2u\sin \theta }{g}$=10

$T_2=\frac{2u\sin \left(90°-\theta \right)}{g}=\frac{2u\cos \theta }{g}=5$

So,

$u\sin \theta =\frac{10g}{2}=50$

$u\cos \theta =\frac{5g}{2}=25$

Range:

$R=\frac{2\left(u\sin \theta \right)\left(u\cos \theta \right)}{g}$

$R=\frac{2\left(50\right)\left(25\right)}{10}=250m$

$\frac{r_1}{r_2} =\frac{3}{4}$

As centripetal force F $=\frac{m{v}^2}{r}$
In order to have constant (same in this question) centripetal force

\begin{aligned}
& F_1=F_2 \\
& \frac{m_1 v_1^2}{r_1}=\frac{m_2 v_2^2}{r_2} \\
& \Rightarrow \frac{v_1}{v_2}=\sqrt{\frac{r_1}{r_2}}=\frac{\sqrt{3}}{2}
\end{aligned}

1. Initial Vertical Velocity ($u_y$):

   $$u_y = u \sin \theta = 60 \sin 30^\circ = 30 \, \text{m/s}.$$

2. Time of Flight (Total time of upward motion):
   Using the formula for the total time of motion upward:

   $$T_{\text{up}} = \frac{u_y}{g} = \frac{30}{10} = 3 \, \text{seconds}.$$

   The last second in the upward motion corresponds to the time $t = 2 \, \text{seconds} \to 3 \, \text{seconds}$

Using the equation of motion:

$$h = u_y t - \frac{1}{2} g t^2,$$

for $t = 1$:

$$h_1 = (30)(1) - \frac{1}{2}(10)(1)^2 = 30 - 5 = 25 \, \text{m}.$$

For the last second in the upward motion, the time interval is $t = 2 \to 3 \, \text{seconds}$. To find $h_2$, calculate the difference in height between these two time intervals:

$$h_2 = \text{Height at } t = 3 \, \text{seconds} - \text{Height at } t = 2 \, \text{seconds}.$$

$$h(t=2) = (30)(2) - \frac{1}{2}(10)(2)^2 = 60 - 20 = 40 \, \text{m}.$$

$$h(t=3) = (30)(3) - \frac{1}{2}(10)(3)^2 = 90 - 45 = 45 \, \text{m}.$$

$$h_2 = h(t=3) - h(t=2) = 45 - 40 = 5 \, \text{m}.$$

$$\frac{h_1}{h_2} = \frac{25}{5} = 5:1.$$

Final Answer:

$$\boxed{5:1}$$​

$$\begin{gathered} K{E}_{\text{initial }}=\frac{1}{2}m{v}^2 \\ K{E}_{\text{initial }}=\frac{1}{2}\times 0.1\times (20{)}^2=20J \\ K{E}_{\text{final }}=\frac{1}{2}m{v}_x^2 \\ K{E}_{\text{final }}=\frac{1}{2}\times 0.1\times (10{)}^2=5J\text{ Decrease in Kinetic Energy } \\ \Delta KE=K{E}_{\text{initial }}-K{E}_{\text{final }} \\ \Delta KE=20-5=15J \end{gathered}$$

The maximum height of the projectile is given by

$$\begin{gathered} \mathrm{H} = \frac{{\mathrm{u}}^2 {\sin }^2\mathrm{\theta }}{2\mathrm{g}} \\ \mathrm{So}, \\ \frac{{\mathrm{H}}_1}{{\mathrm{H}}_2} = \frac{{\sin }^2{\mathrm{\theta }}_1}{{\sin }^2{\mathrm{\theta }}_2} = \frac{{\sin }^{2}60°}{{\sin }^{2}30°} \\ \frac{{\mathrm{H}}_1}{{\mathrm{H}}_2} =\sqrt{3} : 1 \end{gathered}$$

$H_{\max }₁=8\times H_{\max }₂$

$\frac{u^2{\sin {\theta }_1}^2}{2g}=8\times \frac{u^2{\sin {\theta }_2}^2}{2g}$

$\sin {\theta }_1=2\sqrt{2}\sin {\theta }_2$

$\frac{T_1}{T_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}$

$\frac{T_1}{T_2}=2\sqrt{2}$