To calculate impulse first of all calculate the time during which force becomes zero.
We have given, $$F = 600 - 2 \times {10^5}t$$
When, bullet leaves the barrel, the force on the bullet becomes zero.
$$\eqalign{ & {\text{So,}}\,600 - 2 \times {10^5}t = 0 \cr & \Rightarrow t = \frac{{600}}{{2 \times {{10}^5}}} \cr & = 3 \times {10^{ - 3}}s \cr} $$
Then, average impulse imparted to the bullet
$$\eqalign{ & I = \int_0^t F dt = \int_0^{3 \times {{10}^{ - 3}}} {\left( {600 - 2 \times {{10}^5}t} \right)} dt \cr & = \left[ {600t - \frac{{2 \times {{10}^5}{t^2}}}{2}} \right]_0^{3 \times {{10}^{ - 3}}} \cr & = 600 \times 3 \times {10^{ - 3}} - {10^5} \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} \cr & = 1.8 - 0.9 = 0.9\;N - s \cr} $$
Alternative
As obtained in previous method, the time taken by bullet when it leaves the barrel
$$t = 3 \times {10^{ - 3}}\;s$$
Let $${F_1}$$ and $${F_2}$$ denote the forces at the time of firing of bullets i.e. at $$t = 0$$  and at the time of leaving the bullet i.e. at $$t = 3 \times {10^{ - 3}}\;s.$$
$$\eqalign{ & {F_1} = 600 - 2 \times {10^5} \times 0 = 600\;N \cr & {F_2} = 600 - 2 \times {10^5} \times 3 \times {10^{ - 3}} = 0 \cr} $$
Mean value of force $$F = \frac{1}{2}\left( {{F_1} + {F_2}} \right) = \frac{{600 + 0}}{2} = 300\;N$$
Thus, impulse $$ = F \times t = 300 \times 3 \times {10^{ - 3}} = 0.9\;N - s$$

Given: Mass of rocket $$\left( m \right) = 5000\,kg$$
Exhaust speed $$\left( v \right) = 800\,m/s$$
Acceleration of rocket $$\left( a \right) = 20\,m/{s^2}$$
Gravitational acceleration $$\left( g \right) = 10\,m/{s^2}$$
We know that upward force
$$F = m\left( {g + a} \right) = 5000\left( {10 + 20} \right) = 5000 \times 30 = 150000\,N.$$
We also know that amount of gas ejected
$$\left( {\frac{{dm}}{{dt}}} \right) = \frac{F}{v} = \frac{{150000}}{{800}} = 187.5\,kg/{s^{ - 1}}$$

Impulse of a force, which is the product of average force during impact and the time for, which the impact lasts is measured by the total change in linear momentum produced during the impact.
Impulse $$I = {F_{{\text{av}}}} \times t = {P_2} - {P_1}$$
Here, Mass $$ = 150g = \frac{{150}}{{1000}}kg$$
$$\eqalign{ & \therefore F = \frac{{150}}{{1000}} \times 20 = 3\,N \cr & \therefore I = F \cdot \Delta t = 3 \times 0.1 = 0.3\,N - s \cr} $$

From question, Mass of body, $$m = 5\,kg$$
Velocity at $$t = 0,u = \left( {6\hat i - 2\hat j} \right)m/s$$
Velocity at $$t = 10s,v = + 6\hat j\,m/s$$
Force, $$F = ?$$
$$a = \frac{{v - u}}{t} = \frac{{6\hat j - \left( {6\hat i - 2\hat j} \right)}}{{10}} = \frac{{ - 3\hat i + 4\hat j}}{5}\,m/{s^2}$$
Force, $$F = ma = 5 \times \frac{{\left( { - 3\hat i + 4\hat j} \right)}}{5} = \left( { - 3\hat i + 4\hat j} \right)N$$

$$\eqalign{ & {\text{Here}}\,m = 0.5\,kg;u = - 10\,m/s; \cr & t = \frac{1}{{50}}\,s;v = + 15\,m{s^{ - 1}} \cr & {\text{Force}} = \frac{{m\left( {v - u} \right)}}{t} = 0.5\left( {10 + 15} \right) \times 50 = 625\,N \cr} $$

Thrust on the rocket is the force with which the rocket moves upwards. Thrust on rocket at time $$t$$ is given by $$F = - u\frac{{dm}}{{dt}}$$
The negative sign indicates that thrust on the rocket is in a direction opposite to the direction of escaping gases.
Here, velocity of the rocket $$u = 300\,m/s$$
and force $$F = 345\,N$$
∴ Rate of combustion of fuel $$ - \left( {\frac{{dm}}{{dt}}} \right) = \frac{F}{u} = \frac{{345}}{{300}} = 1.15\,kg/s$$

Force imparted = rate of change of momentum
$$\eqalign{ & F = \frac{{\Delta p}}{{\Delta t}} \cr & {\text{or}}\,\,F = \frac{{{p_1} - {p_2}}}{{\Delta t}} \cr & {\text{or}}\,\,F = \frac{{m\left( {{v_1} - {t_2}} \right)}}{{\Delta t}} \cr} $$
Here, mass of body $$m = 150\;g = 0.150\;kg,$$
$$\eqalign{ & {v_1} = 20\;m/s, \cr & {v_2} = 0 \cr} $$
Time taken, $$\Delta t = 0.1\;s$$
$$\eqalign{ & \therefore F = \frac{{0.150 \times \left( {20 - 0} \right)}}{{0.1}} \cr & = 30\;N \cr} $$

As the ball, $$m = 10\,g = 0.01\,kg$$    rebounds after striking the wall
$$\therefore $$ Change in momentum $$ = mv - \left( { - mv} \right) = 2\,mv$$
Impulse = Change in momentum $$= 2mv$$
$$\therefore v = \frac{{{\text{Impulse}}}}{{2m}} = \frac{{0.54\,Ns}}{{2 \times 0.01\,kg}} = 27\,m{s^{ - 1}}$$

$${\text{Given}}\,F = 600 - \left( {2 \times {{10}^5}t} \right)$$
The force is zero at time $$t,$$ given by $$0 = 600 - 2 \times {10^5}t$$
$$\eqalign{ & \Rightarrow t = \frac{{600}}{{2 \times {{10}^5}}} = 3 \times {10^{ - 3}}\,{\text{seconds}} \cr & \therefore {\text{Impulse}} = \int\limits_0^t {Fdt} = \int\limits_0^{3 \times {{10}^{ - 3}}} {\left( {600 - 2 \times {{10}^5}t} \right)} dt \cr & = \left[ {600t - \frac{{2 \times {{10}^5}{t^2}}}{2}} \right]_0^{3 \times {{10}^{ - 3}}} \cr & = 600 \times 3 \times {10^{ - 3}} - {10^5}{\left( {3 \times {{10}^{ - 3}}} \right)^2} \cr & = 1.8 - 0.9 = 0.9\,Ns \cr} $$

As, $$mr = {\text{constant}}$$
$$\eqalign{ & {m_1}{r_1} = {m_2}{r_2} \Rightarrow {r_2} = \frac{{{m_1}{r_1}}}{{{m_2}}} \cr & = \frac{{0.5 \times 10}}{{50}} = 0.1 \cr} $$
So, distance travelled by the man will be $$10 + 0.1 = 10.1\,m$$