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David G

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This is obviously assuming no air resistance.

The bullet takes 130/200 = 13/20 s to reach the target. In that time it has been acted on by gravity. Initial vertical velocity = 0. Acceleration = g = 10 m/s² (roughly)
drop = g t²/2 = 169/80 ~ 2.11 m (which is why you don't engage targets at 200m with such a slow bullet)



We need to impart sufficient upward velocity to the bullet such that at 65m its vertical velocity is zero. If the elevation is α, then the initial vertical velocity is 200 sin α and its initial horizontal velocity is 200 cos α

It will take t = 65/(200 cos α) to reach midpoint of trajectory. At this point its vertical velocity will be 200 sin α - g t. Equate to zero.

Answers are ~ 0.93° and 89.1°

There are always two answers because the underlying solution depends on a quadratic equation.
 
A hunter aims directly at a target (on the same level) 130 m away.

(a) If the bullet leaves the gun at a speed of 200 m/s, by how much will it miss the target?

(b) At what angle should the gun be aimed so as to hit the target?
 
This is an exercise in kinematics. We start with: how much will the bullet accelerate before it reaches to target?

I'm going to assume you're shooting in a vacuum. A perfectly reasonable assumption, right? In that case, only gravity is doing the accelerating. To know how much the bullet accelerates, we need to know how long it's accelerating.

It takes the bullet (130 m) / (200 m/s) = 0.65 seconds to reach the target. Determine the distance the bullet moves due to gravitational acceleration to determine how much is misses by.

Assuming you're shooting in a vacuum on Earth (I do it all the time): Accelerating at 9.81 m/s^2, the bullet moves 1/2 * (-9.81 m/s^2) * (0.65 s)^2 = 2.072 m. The bullet will be off by -2.072 m.

To determine the angle you should fire at, realize you have a few equations.

Now, notice that if you're firing at an angle theta:
s_y(t) = 1/2 * -9.81 m/s^2 * t^2 + 200 m/s * sin(theta) * t
s_x(t) = 200 m/s * cos(theta) * t

We want the bullet to hit the target, which means we want s_x(t_i) = 130 m and s_y(t_i) = 0, where t_i is the time of impact. We can solve for the time of impact (using s_x(t_i) = 130 m) by:

130 m = 200 m/s * cos(theta) * t_i
t_i = 0.65 s * cos(theta)

If we plug that back into s_y(t_i) we get:
0 = 1/2 * -9.81 m/s^2 * (0.65 s * cos(theta))^2 + 200 m/s * sin(theta) * 0.65 s * cos(theta)

Solving for theta we get 0.015939899857365027 radians and 1.5707963267948966 radians. If we use the correct number of significant figures, it's just 0.02 (the velocity is only specified to one significant figure).

How did I solve for theta? I took a shortcut: http://www.wolframalpha.com/input/?i=0+%3D+1%2F2+*+-9.81+m%2Fs^2+*+(0.65+s+*+cos(theta))^2+%2B++200+m%2Fs+*+sin(theta)+*+0.65+s+*+cos(theta)

If you need to show your work on that, try using a list of trigonometric identities.
 
D=(1/2)(acceleration)(time)^2 + Vit(initial velocity multiplied by time).

130=(1/2)(9.8m/s^2)t^2 + (200m/s)t

Use quadratic formula; t=6.4*10^-1s

Next...

total distance = initial distance + (initial velocity)(time) +(1/2)(acceleration)(time)^2

x= 130m + (0m/s)(.54s) + (1/2)(9.8m/s^2)(.64s)
x=1.33*10^2m

The hunter is off by about 3 some odd meters.
 
This is obviously assuming no air resistance.

The bullet takes 130/200 = 13/20 s to reach the target. In that time it has been acted on by gravity. Initial vertical velocity = 0. Acceleration = g = 10 m/s² (roughly)
drop = g t²/2 = 169/80 ~ 2.11 m (which is why you don't engage targets at 200m with such a slow bullet)



We need to impart sufficient upward velocity to the bullet such that at 65m its vertical velocity is zero. If the elevation is α, then the initial vertical velocity is 200 sin α and its initial horizontal velocity is 200 cos α

It will take t = 65/(200 cos α) to reach midpoint of trajectory. At this point its vertical velocity will be 200 sin α - g t. Equate to zero.

Answers are ~ 0.93° and 89.1°

There are always two answers because the underlying solution depends on a quadratic equation.
 
the bullet will drop 32 feet per sec squared.

130 m / 200 m/s = number of seconds bullet is in air.

that number squared times 32 feet/sec^2 gives the feet the bullet will drop and miss the target.

if you know the rise (the drop) and the run (130 m) you can calculate the angle to raise the gun
 
D=(1/2)(acceleration)(time)^2 + Vit(initial velocity multiplied by time).

130=(1/2)(9.8m/s^2)t^2 + (200m/s)t

Use quadratic formula; t=6.4*10^-1s

Next...

total distance = initial distance + (initial velocity)(time) +(1/2)(acceleration)(time)^2

x= 130m + (0m/s)(.54s) + (1/2)(9.8m/s^2)(.64s)
x=1.33*10^2m

The hunter is off by about 3 some odd meters.
 
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