# Projectile motion of a moving target

• shibe
In summary, the conversation discusses problem #57 from chapter 4 of "Physics for engineers and scientists vol1" and the different methods used to solve it. The author's method of successive approximation is described, where they first calculate the flight time of the projectile and then correct for the ship's displacement. However, the speaker wanted to directly describe the motion of both the ship and the bullet and then solve for the time at which their displacements were equal. They attempted to do this by defining unit vectors and creating displacement functions, but encountered issues with the bullet's velocity and gravity. They are seeking help in constructing displacement functions or advice on whether to stick with the author's method.
shibe
Homework Statement
A ship is steaming at 30 km/h on a course parallel to a straight
shore at a distance of 17000 m. A gun emplaced on the shore
(at sea level) fires a shot with a muzzle speed of 700 m/s when
the ship is at the point of closest approach. If the shot is to hit
the ship, what must be the elevation angle of the gun? How
far ahead of the ship must the gun be aimed?
Relevant Equations
v=at
this is problem #57 from chapter 4 of "Physics for engineers and scientists vol1" ,it was solved in the book by a method the author calls "successive approximation, he first calculated the flight time of the projectile and then "corrected" for the ship's displacement. i know this is a perfectly good method, but i wanted to directly describe the motion of both the ship and the bullet (relative to the bullet's point of projection) and then solve for t at which their displacements were equal. this initial attempt resulted in:
ship:
$$x(t)= \vec i(17000)+\vec j(0) + \vec k(\frac{25t}{3})$$
bullet:
$$x(t)=\vec i(700\cos\theta)+\vec j(700\sin\theta)+\vec k(0)$$

where I,J,K were defined as the unit vectors in the direction of the ship from the gun, the direction upward, and the direction in which the ship travels respectively.

But this had to be incorrect, since the projected bullet actually travels in a 2D plane connecting the origin and its point of impact(with the ship) such that the resolution of the bullet's velocity in that plane does not yield vectors that were perpendicular to the velocity of the ship (which i previously termed k). This is the main obstacle to my much desired displacement(t) functions.

i have tried to make the plane in which the bullet travels and its relation to the ship's velocity vector clearer with a pathetically(sorry) drawn picture.
hopefully someone can help in the construction of displacement(t) functions with a decent explanation i can follow. or maybe deliver the bad news explaining why its unreasonable and i should just stick with the author's method. thanks

#### Attachments

• Projectile motion #57.png
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Several things wrong with your shell (bullet?) equation.
• You have a position on the left but velocities on the right.
• You need to allow a k component of launch velocity.
• You forgot gravity.

## 1. What is projectile motion?

Projectile motion is the motion of an object that is projected into the air at an angle and then moves along a curved path under the influence of gravity.

## 2. How does the motion of a moving target affect projectile motion?

The motion of a moving target does not affect the projectile motion itself. However, it may add an additional velocity component to the initial velocity, which can change the trajectory of the projectile.

## 3. What factors affect the trajectory of a projectile?

The trajectory of a projectile is affected by the initial velocity, angle of projection, air resistance, and the force of gravity.

## 4. How can the distance and time of flight of a projectile be calculated?

The distance of a projectile can be calculated using the formula d = v0t + 1/2at^2, where d is distance, v0 is initial velocity, a is acceleration due to gravity, and t is time. The time of flight can be calculated using the formula t = 2v0sinθ/g, where θ is the angle of projection and g is the acceleration due to gravity.

## 5. Can a projectile reach the same height at its maximum point as it was projected from?

Yes, a projectile can reach the same height at its maximum point as it was projected from if it is projected at a 45-degree angle and there is no air resistance. This is known as the range equaling the maximum height in projectile motion.

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