Answer:
1,000 N
Explanation:

Force exerted by the bullet on the rifle = 
F = 1000N
Contextual Way:
Newton's second law of motion.
F=m a
Now to solve based on the current info, we shall assume that:-
The force exerted on the bullet was uniform across the entire duration of bullet leaving the barrel, i.e., 0.003 seconds {Not necessarily true for real life applications as the force will not be uniform from the point of hammer impact till the point of leaving the barrel. In reality you will get a Force profile across that entire duration}
We are not distinguishing between bullet and cartridge. {What you shall hold in your hand and load in a revolver is a cartridge containing the gunpowder, bullet etc. The bullet is the projectile at the mouth of the cartridge that actually leaves the barrel and hit the target. So when you are weighing in real life, you are not weighing the bullet, rather the cartridge as a whole}
Getting back to the question
Impulse equation for the bullet
∫F∗dt=∫m∗dv
Average impulse delivered= Change in momentum of the bullet
Assuming average force delivery
Favg∗∫dt=m∗∫dv
Favg∗0.003=0.010∗300
Favg=300∗10/3
Favg=1000N
Work = Force x Distance = 500 x 4 = 2000 Nm = 2000 J
Answer:
Part a)

Part b)

Since the distance of other building is 15 m so YES it can make it to other building
Part c)

direction of velocity is given as
![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Explanation:
Part a)
acceleration due to gravity on this planet is 3/4 times the gravity on earth
So the acceleration due to gravity on this new planet is given as


now the vertical displacement covered by the canister is given as

now by kinematics we have



Part b)
Horizontal speed of the canister is given as

now the distance moved by it



Since the distance of other building is 15 m so YES it can make it to other building
Part c)
Final velocity in X direction will remains the same

final velocity in Y direction



now magnitude of velocity is given as



direction of velocity is given as


![[tex]\theta = 26.35 degree](https://tex.z-dn.net/?f=%5Btex%5D%5Ctheta%20%3D%2026.35%20degree)
Answer: a) 19.21m b) 3.92secs
Explanation:
a) Maximum height reached by the object is the height reached by an object before falling freely under gravity.
Maximum height = U²/2g
U is the initial velocity = 19.6m/s
g is acceleration due to gravity = 10m/s²
Maximum Height = 19.6²/2(10)
H = 19.21m
b) The time elapsed before the stone hits the ground is the time of flight T= 2U/g
T= 2(19.6)/10
T = 39.2/10
Time elapsed is 3.92secs