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Step-by-Step Solution
Step 1: Identify the Given Data
• Mass of the body, $m = 10\,\text{kg}$
• Angle of projection, $\theta = 45^\circ$
• A point on the trajectory: $\bigl( x = 20\,\text{m},\,y = 10\,\text{m}\bigr)$
• Acceleration due to gravity, $g = 10\,\text{m/s}^2$
Step 2: Understand the Relationship for a Projectile's Trajectory
For a projectile fired at an angle $\theta$ with initial speed $u$, the standard form of the trajectory is:
$y = x \tan \theta \left(1 - \frac{x}{R}\right)$,
where $R$ is the horizontal range given by
$R = \frac{u^2 \sin 2\theta}{g}$.
Step 3: Use the Given Point to Find the Range
We know the point $(20, 10)$ lies on the trajectory. Thus,
$10 = 20 \,\tan 45^\circ \left(1 - \frac{20}{R}\right).$
Since $\tan 45^\circ = 1$, this becomes:
$10 = 20 \left(1 - \frac{20}{R}\right).$
Divide both sides by 20:
$\frac{10}{20} = 1 - \frac{20}{R}, \quad \text{or}\quad 0.5 = 1 - \frac{20}{R}.$
Hence,
$\frac{20}{R} = 1 - 0.5 = 0.5 \quad\Longrightarrow\quad R = 40\,\text{m}.$
Step 4: Determine the Initial Speed
Using the range formula for a projectile,
$R = \frac{u^2 \sin 2\theta}{g}.$
Given $R = 40\,\text{m}$, $\theta = 45^\circ$ (so $\sin 90^\circ = 1$), and $g = 10\,\text{m/s}^2$:
$40 = \frac{u^2 \times 1}{10} \quad\Longrightarrow\quad u^2 = 400 \quad\Longrightarrow\quad u = 20\,\text{m/s}.$
Step 5: Compute the Total Time of Flight
The total time of flight $T$ for a projectile launched at speed $u$ and angle $\theta$ is:
$T = \frac{2u \sin \theta}{g}.$
Since $u = 20\,\text{m/s}$ and $\sin 45^\circ = \frac{1}{\sqrt{2}}$,
$T = \frac{2 \times 20 \times \frac{1}{\sqrt{2}}}{10} \;=\; \frac{40}{10\sqrt{2}} \;=\; \frac{4}{\sqrt{2}}\,\text{s}.$
Step 6: Find the Instant for Which Momentum is Required
We are interested in the time $\displaystyle t = \frac{T}{\sqrt{2}}$. Given $T = \frac{4}{\sqrt{2}}\,\text{s}$,
$t = \frac{1}{\sqrt{2}}\times \frac{4}{\sqrt{2}} = \frac{4}{2} = 2\,\text{s}.$
Step 7: Calculate the Velocity at $t=2\,\text{s}$
The horizontal velocity component remains constant throughout the flight (no horizontal acceleration):
$v_x = u \cos \theta = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} \,\text{m/s}.$
The vertical velocity component changes due to gravity. At any time $t$:
$v_y = u \sin \theta - gt.$
Here, $u \sin 45^\circ = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}$, so at $t = 2\,\text{s}$:
$v_y = 10\sqrt{2} - (10 \times 2) = 10\sqrt{2} - 20\,\text{m/s}.$
Step 8: Determine the Momentum
Momentum $\overrightarrow{p}$ is given by:
$\overrightarrow{p} = m\,\overrightarrow{v} = m\,\bigl(v_x\,\hat{i} + v_y\,\hat{j}\bigr).$
Substituting $m = 10\,\text{kg}$, $v_x = 10\sqrt{2}$, and $v_y = 10\sqrt{2} - 20$:
$\overrightarrow{p} = 10 \left[ 10\sqrt{2}\,\hat{i} + \bigl(10\sqrt{2} - 20\bigr)\,\hat{j} \right].$
Hence,
$\overrightarrow{p} = 100\sqrt{2}\,\hat{i} + \bigl(100\sqrt{2} - 200\bigr)\,\hat{j}.$
Final Answer
$ \boxed{ \displaystyle 100 \sqrt{2}\,\hat{i} \;+\;\bigl(100 \sqrt{2} - 200\bigr)\,\hat{j} } $
