Projectile Motion Calculations: Ship's Shot at an Enemy Ship and Island Peak

AI Thread Summary
A ship fires a projectile at an enemy ship while maneuvering near an island's mountain peak, seeking to determine how close the projectile lands to the enemy ship and its vertical proximity to the peak. The projectile is launched with an initial velocity of 250 m/s at a 75-degree angle. Key equations for vertical and horizontal motion are discussed, including the effects of gravity on the projectile's trajectory. Participants encourage the original poster to show their calculations and utilize available resources for further guidance. Understanding the physics principles involved is essential for solving the problem accurately.
lucianman24
Messages
1
Reaction score
0
Please help me with the following question:

A ship maneuvers to within 2.50 x 10 ^ 3 m of an island's 1.80 x 10 ^ 3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10 ^ 2 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 2.50 x 10 ^ 2 m/s at an angle of 75 degrees, how close to the enemy ship does the projectile land? How close (vertically) does the projectile come to the peak? My physics book says the answer will be 8m and 210m but I want to know how to do it.

Vertical Motion of a Projectile that falls from rest
Vy,f = -gt (where g = a = 9.81) (t = time)
Vy,f^2 = -2gy (y = delta y= displacement in the y direction)
delta y = -1/2g(t)^2

Horizontal Motion of a Projectile
Vx = Vx,i = constant ( i = intial velocity)
delta x = Vxt ( delta x = dispalcement in the x direction)

Projectiles Launched At An Angle
Vx = Vi (cos theta) = constant (theta = degrees of angle)
delta x = Vi (cos theta)t
Vy,f = Vi (sin theta) - gt
Vy,f^2 = Vi^2 (sin theta)^2 - 2g(delta y)
delta y = Vi (sin theta)t - 1/2g(t)^2

Hope you can understand if not reply saying what you dont
Thank You
 
Last edited:
Physics news on Phys.org
lucianman24 said:
Please help me with the following question:

A ship maneuvers to within 2.50 x 10 ^ 3 m of an island's 1.80 x 10 ^ 3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10 ^ 2 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 2.50 x 10 ^ 2 m/s at an angle of 75 degrees, how close to the enemy ship does the projectile land? How close (vertically) does the projectile come to the peak?


THANK YOU

Hey, lucianman, welcome to the site!

What's your work so far for this problem? What values do you know? What equations might be useful?
 
lucianman24 said:
Please help me with the following question:

A ship maneuvers to within 2.50 x 10 ^ 3 m of an island's 1.80 x 10 ^ 3 m high mountain peak and fires a projectile at an enemy ship 6.10 x 10 ^ 2 m on the other side of the peak. If the ship shoots the projectile with an initial velocity of 2.50 x 10 ^ 2 m/s at an angle of 75 degrees, how close to the enemy ship does the projectile land? How close (vertically) does the projectile come to the peak?

THANK YOU

Show us some work. You should start off by writing down all the equations related to projectile motion (i.e. velocity, displacement, etc.).

P.S. Typing 'projectile motion' into the search box should be very useful too, since it is a highly frequent topic. :smile:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top