I have been struggling with this problem for two days. I tried to apply the basic kinematic equations to the problem but couldn't figure it out© BrainMass Inc. brainmass.com December 24, 2021, 4:53 pm ad1c9bdddf
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A high-powered rifle fires a bullet with a muzzle speed of 1.00 km/s. The gun is pointed horizontally at a large bull's eye target - a set of concentric rings - 200 m away.
(a) How far below the extended axis of the rifle barrel does a bullet hit the target?
Since the horizontal and vertical components of the velocity of a projectile behave independent of each other, we use the following two equations to describe those components:
Eq 1 vx = v0x
where v0x is the initial horizontal velocity and vx is the horizontal velocity at any time. (Notice that the horizontal velocity does not depend on time.)
Eq 2 vy = v0y + g t
Where v0y is the initial vertical velocity and vy is the vertical velocity at any time. (Notice that the vertical velocity does depend on time since g = -9.8 m/s2)
We have v0x = 1km/s = 1000 m/s
One can also find relationships for the distance travelled in each direction. Equations giving these distances are as follows:
Eq 3 x = v0x t Gives the horizontal distance travelled
Eq 4 y = v0y t + (1/2) g t2 Gives the vertical distance travelled
From eq 3, we get time of flight = x/v0x = 200/1000 = 0.2 Sec
Now using eq. 4, y = v0y * t + (1/2) g t2 = 0 + (1/2) g t2
= 0.5 * 9.8 * 0.22 = 0.196m (Ans)
The rifle is equipped with a telescopic sight. It is "sighted in" by adjusting the axis of the telescopes that it points precisely where the bullet hits the target at 200 m.
(b) Find the angle between the telescope axis and the rifle barrel axis.
Find the vertical (v0y) and horizontal (v0x) components of the muzzle velocity from:
v0y = v0 sin (q) and v0x = v0 cos (q) where V (0) is the initial velocity and q is the angle at which it will be fired.
Here we have compensated the 0.196m by adjusting the barrel at an angle. (if the gun is at an angle q above the horizontal, it will go down 0.196m near the target and will hit it directly)
Again we can write the vertical displacement, y = v0y t + (1/2) g t2
But here v0y = v0 sin (q)
Time of flight t = 2 v0 sin (q) /g = [2 * 1000/9.8] Sin q = 204.08 Sin q
y = v0y t + (1/2) g t2
 0.196 = v0 sin (q) [204.08 Sin q] + 0.5 * 9.8 * [204.08 Sin q] 2
 0.196 = 204.8*1000 Sin2q + 204078.367 Sin2q
From this sin2q = 0.196/408878.367 = 4.79*10-7
sin q = 6.9*10-4  q = 0.039 deg
When shooting at a target at a distance other than 200 m, the marksman uses the telescopic sight, placing the crosshairs to "aim high" of "aim low" to compensate for the different range. Should she aim high or low, and approximately how far from the bull's eye, when the target is at a distance of
(c) 50.0 m,
(d) 150.0 m, or
(e) 250.0 m?
c) Time of flight = 50/1000 = 0.05 sec
y = v0y t + (1/2) g t2 = 0 + 4.9 *0.052 = 0.01225 m = 1.225 cm
d) Time of flight = 150/1000 = 0.15 sec
y = v0y t + (1/2) g t2 = 0 + 4.9 *0.152 = 0.11025 m = 11.025 cm
e) Time of flight = 250/1000 = 0.25 s
y = 4.9*0.252 = 0.30625 m = 30.625 cm
These are the vertical distance the bullet will travel even when it is fired horizontally. We must compensate this distance by adjusting the angle of the gun.
Note: The trajectory of the bullet is everywhere so nearly horizontal that it is a good approximation to model the bullet as fired horizontally in each case. What if the target is uphill or downhill?
(f) Suppose the target is 200 m away, but the sight line to the target is above the horizontal by 30.0 degrees. Should the marksman aim high, low, or right on?
If you are taking the trajectory be a straight line, the marksman must point the gun high with respect to the horizontal.
(g) Suppose the target is downhill by 30.0 degrees. Should the marksman aim high, low, or right on?
If you are taking the trajectory be a straight line, the marksman must point the gun low with respect to the horizontal
If the bullet is not a projectile, it will follow a straight line. In the case of a projectile, the trajectory will be a parabola.