Correct option is (d)

To find the angle of projection for a projectile to have the same horizontal range and maximum height, we need to express both the range and the maximum height in terms of the projectile's initial velocity and the angle of projection, and then set them equal to each other.

The formula for the horizontal range $R$ of a projectile is given by:

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

where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $\theta$ is the angle of projection.

The formula for the maximum height $H$ reached by the projectile is:

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

To have the same numerical value for $R$ and $H$, we set them equal to each other:

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

Simplifying this equation, we get:

$2\sin 2\theta ={\sin }^2\theta .$

Using the double-angle formula, $\sin 2\theta =2\sin \theta \cos \theta$, we can rewrite the equation as:

$4\sin \theta \cos \theta ={\sin }^2\theta .$

This can be simplified further to:

$4\sin \theta \cos \theta =(\sin \theta {)}^2.$

Dividing both sides by $\sin \theta$ (assuming $\sin \theta \ne 0$), we get:

$4\cos \theta =\sin \theta .$

Now, dividing both sides by $\cos \theta$, we have:

$4=\tan \theta .$

So, the angle of projection $\theta$ is:

$\theta ={\tan }^{-1}(4).$

Therefore, the correct answer is:

Option D ${\tan }^{-1}(4)$.

Correct option is ()

For ${\mathrm{u}}_{\mathrm{A}}$ & ${\mathrm{u}}_{\mathrm{B}}$ time of flight and range can not be same. So above options are incorrect.

Correct answer is 100

To solve this problem, we can use the equations of motion under uniform acceleration, separately considering the horizontal and vertical motions because the two are independent of each other.

First, for the body of mass $M$ thrown horizontally with velocity $v$ from a height $H$, let's analyze its motion:

rizontal Motion:

The horizontal distance (range) $x$ covered by the object is given by $x=v\cdot t$, where $t$ is the time taken to hit the ground.

rtical Motion:

The time $t$ it takes for the object to hit the ground can be found using the equation of motion under gravity, $H=\frac{1}{2}g{t}^2$, where $g$ is the acceleration due to gravity.

r the first body:

Given $x=100$ m and using the equation for the vertical motion to find $t$, we have:

$H=\frac{1}{2}g{t}^2$

Solving for $t$, we get:

$t=\sqrt{\frac{2H}{g}}$

The horizontal motion gives:

$x=v\cdot t\Rightarrow 100=v\cdot \sqrt{\frac{2H}{g}}$

Now, considering the second body of mass $2M$ thrown at velocity $\frac{v}{2}$ from height $4H$:

r the second body:

The time $t^{\prime }$ it takes for the second body to hit the ground from height $4H$ can be found by:

$4H=\frac{1}{2}g{t}^{\prime 2}$

Solving for $t^{\prime }$, we get:

$t^{\prime }=\sqrt{\frac{2\cdot 4H}{g}}=2\sqrt{\frac{2H}{g}}$

This is twice the time $t$ found for the first body.

The horizontal distance $x^{\prime }$ covered by the second body is:

$x^{\prime }=\left(\frac{v}{2}\right)\cdot t^{\prime }$

Now, substituting the value of $t^{\prime }$ found above, we get:

$x^{\prime }=\left(\frac{v}{2}\right)\cdot 2\sqrt{\frac{2H}{g}}=v\cdot \sqrt{\frac{2H}{g}}$

But we previously found that $v\cdot \sqrt{\frac{2H}{g}}=100$, so:

$x^{\prime }=100\text{ m}$

Therefore, a body of mass $2M$ thrown horizontally with velocity $v/2$ from the top of a tower of height $4H$ will touch the ground at a distance of 100 meters from the foot of the tower.

Correct answer is 16

To solve this problem, we first need to understand the formula that relates the maximum height $H$ reached by a projectile to its initial velocity $v_0$ and the acceleration due to gravity $g$:

$H=\frac{v_0^2{\sin }^2(\theta )}{2g}$

where:

• $H$ is the maximum height,


• $v_0$ is the initial velocity of the projectile,


• $\theta$ is the angle of projection with the horizontal, and


• $g$ is the acceleration due to gravity, which is approximately $9.8\mathrm{m}/{\mathrm{s}}^2$.

Given that the maximum height attained by the projectile is $64\mathrm{m}$, we can write:

$64=\frac{v_0^2{\sin }^2(\theta )}{2\cdot 9.8}$

Now, we are asked to find the new maximum height if the initial velocity is halved. Let's denote the new initial velocity as $v_0^{\prime }=\frac{v_0}{2}$. Using the formula for maximum height again, we get:

$H^{\prime }=\frac{{v_0^{\prime }}^2{\sin }^2(\theta )}{2g}$

Substituting $v_0^{\prime }=\frac{v_0}{2}$ into this equation:

$H^{\prime }=\frac{(\frac{v_0}{2}{)}^2{\sin }^2(\theta )}{2g}=\frac{v_0^2{\sin }^2(\theta )}{2\cdot 4g}=\frac{1}{4}\cdot \frac{v_0^2{\sin }^2(\theta )}{2g}$

Since we know the original height:

$64=\frac{v_0^2{\sin }^2(\theta )}{2\cdot 9.8}$

Substituting this value into our equation for $H^{\prime }$, we find:

$H^{\prime }=\frac{1}{4}\cdot 64=16\mathrm{m}$

Therefore, if the initial velocity of the projectile is halved, the new maximum height reached by the projectile would be $16\mathrm{m}$.

Correct answer is 2

The ball needs to just cross 4 steps to just hit $5^{\text{th }}$ step

Therefore, horizontal range $(R)=0.4\mathrm{m}$

$\mathrm{R}=\text{ u.t }$

Similarly, in vertical direction

$\begin{array}{rl}& \mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2\\ & 0.4=\frac{1}{2}{\mathrm{gt}}^2\\ & 0.4=\frac{1}{2}\mathrm{g}{\left(\frac{0.4}{\mathrm{u}}\right)}^2\\ & {\mathrm{u}}^2=2\\ & \mathrm{u}=\sqrt{2}\mathrm{m}/\mathrm{s}\end{array}$

Therefore, $\mathrm{x}=2$