Difficult projectile problem

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In summary, the separation of two points that are both a distance h above the horizontal when a projectile is fired with an initial velocity v and adjusted for maximum range is given by the formula d = v/g \sqrt{v^2 - 4gh}. This can be derived by using the formula for projectile motion and solving for the distance between two points. Another approach is to use conservation of energy and observe that the angle of the projectile is 45 degrees, allowing for the computation of time to reach height h and solving for the distance d.
  • #1
ozone
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A projectile is fired with an initial velocity v such that it passes through two points that are both a distance h above the horizontal. Show that if the gun is adjusted for maximum range the separation of the two points is

[itex] d = v/g \sqrt{v^2 - 4gh} [/itex]

Homework Equations



I have been struggling at this problem for much of the day. I even derived the original equations for range and max height

range, height [itex] R = v^2/g , H = v^2 / 4g [/itex]
Also we have our conservation of motion equations, and the distance function of a projectile (under the constraints of max range)
[itex] F(t) = (1/\sqrt{2} vt) \overline{i} + (1/\sqrt{2} vt - 1/2 gt^2) \overline{j} [/itex]

The Attempt at a Solution


I attempted to use conservation of energy to find a solution to this problem, but I was having a lot of difficulty relating it to my distance function. I also tried using my distance function but I found that I could not effectively remove the time variable.
 
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  • #2
Use the formula for projectile which does not contain t.
y = x tan(theta) - g x^2sec^(theta)/2v^2
Put y = h and solve for x. Difference between x gives you the distance d.
 
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  • #3
Thanks.. putting y in terms of x made this problem quite easy to solve. I wish I had seen it myself.
 
  • #4
Another way to solve it would be to observe that since the angle is 45, you effectively know the vertical and horizontal components of the velocity. So you can immediately compute time t to height h. The equation involved is quadratic, so it gives you two roots. Then the distance sought is the difference between the roots multiplied by the horizontal speed.
 
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After struggling for a while, I decided to try using the given equation for range and substituting it into the equation for max height. This gave me the following equation:

H = R/4

From here, I was able to solve for v in terms of g and H:

v = \sqrt{4gH}

Then, I substituted this value for v into the equation for range:

R = (4gH)/g = 4H

Next, I used the equation for distance to find the time it takes for the projectile to reach the second point:

d = (1/\sqrt{2} vt) \overline{i} + (1/\sqrt{2} vt - 1/2 gt^2) \overline{j}

Solving for t, I got:

t = \sqrt{2d/v}

Finally, I substituted this value for t into the equation for range and solved for d:

R = v^2/g = (4gH)/(g) = 4H

d = \sqrt{2(4H)/v} = 2\sqrt{2Hv}

Therefore, the separation of the two points is:

d = 2\sqrt{2Hv} = 2\sqrt{2(\sqrt{4gH})(\sqrt{4gH})} = 2\sqrt{8gH} = 2\sqrt{8gh}

This confirms the given equation:

d = v/g \sqrt{v^2 - 4gh}

Therefore, if the gun is adjusted for maximum range, the separation of the two points is given by this equation.
 

What is a "Difficult Projectile Problem"?

A "Difficult Projectile Problem" is a type of physics problem that involves calculating the trajectory and motion of a projectile, such as a bullet or a ball, in a complex or non-ideal scenario. This could include factors such as air resistance, wind, or changing gravitational forces.

Why are "Difficult Projectile Problems" important in science?

Difficult projectile problems are important because they allow scientists to understand and analyze the behavior of objects in motion in real-world situations. They also help to develop and improve mathematical models that can be used to predict and solve problems in fields such as engineering, physics, and astronomy.

What factors can make a projectile problem difficult?

There are many factors that can make a projectile problem difficult, including air resistance, wind, changing gravitational forces, non-uniform surfaces, and the shape and mass of the object itself. These factors can make it challenging to accurately predict the motion and trajectory of the projectile.

How do scientists approach solving difficult projectile problems?

Scientists approach solving difficult projectile problems by first identifying all the relevant factors and forces that are acting on the projectile. They then use mathematical equations and models, such as Newton's laws of motion, to calculate and predict the projectile's motion. Advanced techniques such as computer simulations may also be used to solve these types of problems.

What are some real-world applications of difficult projectile problems?

Difficult projectile problems have many real-world applications, including predicting the trajectory of a rocket or missile, understanding the flight of a baseball or golf ball, and designing vehicles and structures to withstand the forces of impact. These problems are also important in fields such as ballistics, ball sports, and space exploration.

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