Mechanics: Calculating the minimum height for a bullet to launch from to hit a target

[PastaTapestry]

1. The muzzle speed for a Lee-Enfield rifle is 630 meters per second. Suppose you fire this
rifle at a target 700 meters away and at the same level as the rifle.
a - In order to hit the target, you must aim the barrel at a point above the target. How
many meters above the target must you aim? Pretend there is no air resistance.
$$v_0 = 630$$
$$delta\x = 700$$
$$ g = 9.81$$



2. $$v_x = v_0 cos\theta$$
$$v_y = v_0 sin\theta - gt$$
$$y = v_0y t - 1/2gt^2$$




3. The point at which the bullet reaches its maximum height, $t_(stop)$, has a vertical velocity component of 0. As such, $v_y - gt = 0$. Solving for $t_(stop)$ yields $t_(stop) = v_y / g$. Since at its maximum height the bullet has finished 1/2 of its flight, we may assume $t_(flight) = 2(t_(stop)) = 2v_y / g$. Keeping this equation in mind, we then try to find the angle theta for the bullet.

$$x_(max) = (2v_0^2(sin\theta)(cos\theta))/g$$
$$700 = ((630^2)/(g)) (sin2\theta) [Sin\theta (cos\theta) = sin2\theta]$$
$$\theta = (arcsin(gd/v_o^2))/2$$
$$\theta = 0.49568$$

Knowing theta, we can calculate the flight time.

$$t_flight = 2v_y / g = 2(630sin(.49568)/9.81 = 1.111$$

To check my work, I plugged the relevant information back into the equation to find the y value. Since y = 0 when it hits the target, then I should receive a value of 0 with the relevant information.

$$y = v_0y (t) - 1/2gt^2$$
$$y = (630sin(.49568))(1.111) - 1/2(9.81)(1.111^2)$$
$$y = -3.983 x 10^-5$$

But it doesn't come out to zero. Here I used the rounded calculations, but while doing the actual work, I stored the information (using my TI-83) and used the exact values. It's really close to 0, but according to this, it would still miss the target at the time given. I kept stopping originally because the theta value seems WAY too small - you're practically aiming straight forward!

I'd love some help, since I've been working on this problem for what feels like hours. Please walk through it in as much detail as possible, because it feels like I've approached the problem every which way!

 

[azizlwl]

1. The muzzle speed for a Lee-Enfield rifle is 630 meters per second. Suppose you fire this
rifle at a target 700 meters away and at the same level as the rifle.
a - In order to hit the target, you must aim the barrel at a point above the target. How
many meters above the target must you aim? Pretend there is no air resistance.
---------------------
You target is 700m away horizontally and 0m vertically.
Use vertical and horizontal displacement equations- 2 unknowns, t and θ .

Use trig function with calculated θ to find the height above target.

 

[voko]

The angle you computed is correct. The slight discrepancy you have is inevitable. When I use the angle in the radian measure and use the full possible precision the Windows calculator has to offer, the discrepancy in the vertical position is 10^-31, which can be safely ignored. When you use only a few digits in the angle measure and the time, the discrepancy in the vertical position will be roughly of the same order.

The accuracy you get is better than 10^-4 - which is meaningless, because the accuracy in g is only 10^-2.

 

[PastaTapestry]

Thank you both so much! Was able to get it. Just wanted some reassurance. :]

 

[Nickie]

Hi there, it looks like you've done a great job breaking down the problem and using the relevant equations to solve for the minimum height. However, I think there may be a small error in your calculation for theta. Instead of using the equation $$\theta = (arcsin(gd/v_o^2))/2$$, which gives you a value of 0.49568 radians, you should use the equation $$\theta = tan^{-1}(v_0y/v_0x)$$ which gives you a value of 0.498 radians. This may seem like a small difference, but it can have a significant impact on the final result.

Using this value for theta, we can calculate the flight time as follows:

$$t_{flight} = \frac{2v_0sin\theta}{g} = \frac{2(630)sin(0.498)}{9.81} = 1.112$$

This is slightly different from your calculated flight time of 1.111, but it is much closer to zero when we plug it back into the equation for y.

$$y = v_0y(t) - \frac{1}{2}gt^2 = (630sin(0.498))(1.112) - \frac{1}{2}(9.81)(1.112)^2 = -0.00006$$

As you can see, this value is much closer to zero and suggests that the bullet would indeed hit the target at the given distance and height.

I hope this helps clear up any confusion and provides a more accurate solution to the problem. Keep up the good work!