Projectile Motion and Impact: Solving for Elevation Angle and Time Interval

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a gun firing at a moving ship. The original poster seeks to determine the required elevation angle and the time interval for the projectile to hit the ship, given specific parameters such as distance and muzzle velocity.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss breaking down the problem into horizontal and vertical components and using kinematic equations. There are attempts to derive relationships between the variables involved, such as the time of flight and the horizontal distance covered by the ship.

Discussion Status

Several participants are engaging with the problem, offering hints and suggesting methods to approach the equations. There is an ongoing exploration of how to relate the time of flight to the elevation angle and the horizontal distance, with no explicit consensus reached on the final solution.

Contextual Notes

Participants note the complexity of the problem due to the moving target and the need to eliminate variables to solve for the elevation angle. The original poster expresses uncertainty about the setup and seeks guidance.

Jacobpm64
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Homework Statement


A gun on the shore (at sea level) fires a shot at a ship which is heading directly toward the gun at a speed of 40 km/h. At the instant of firing, the distance to the ship is 15,000 m. The muzzle velocity of the shot is 700 m/s. Pretend that there is no air resistance.
(a) What is the required elevation angle for the gun? Assume g = 9.80 m/s^2.
(b) What is the time interval between firing and impact?


Homework Equations



Uhmm, I'm guessing. tflight = [tex]\frac{2v_{0} sin\alpha }{g}[/tex]

Maybe xmax = [tex]\frac{v^{2}_{0}sin2\alpha}{g}[/tex] will also be applicable.

[tex]v_{x} = v_{0x} = v_{0}cos\alpha[/tex]
[tex]v_{z} = v_{0z} - gt = v_{0} sin \alpha - gt[/tex]
[tex]x = v_{0x}t[/tex]
[tex]z = v_{0z}t - \frac{1}{2} gt^2[/tex]



The Attempt at a Solution


I'm not sure what to do with this problem. Any way I set it up, I end up with more variables than I can solve for.

I would appreciate a hint to throw me in the right direction. Thanks in advance.
 
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You seem to be throwing in equation willy-nilly. Draw a diagram, and do the usual thing of splitting up into horizontal and vertical components, and use the kinematic equations. Your horizontal target will be moving, so x_max=1500-40t (original distance minus the distance the ship covers in the time the bullet is in the air). You should be able to write an equation for the x component of the displacement of the bullet. Eliminate t using an equation for the y component of the displacement of the bullet.
 
[tex]40 \frac{km}{h} * 1000 \frac{m}{km} * \frac{1}{3600} \frac{h}{s} = 11.1 \frac{m}{s}[/tex]

[tex]x_{max} = 15,000 - 11.1 t[/tex]

[tex]x(t) = 700 cos \alpha t[/tex]

[tex]700 cos \alpha t = 15000 - 11.1 t[/tex]

[tex]z(t) = \frac{-1}{2} (9.80) t^2 + 700sin \alpha t[/tex]

I don't know what to do now.. But I'm pretty sure my equations are good.
 
Ok, well use the fact that the displacement, z, is equal to zero when the bullet hits the ship. Can you then solve this equation for t? (Hint: quadratic formula)
 
[tex]0 = \frac{-1}{2} (9.80) t^2 + 700 sin \alpha t[/tex]

By the quadratic equation,

[tex]t = 142.86 sin \alpha[/tex]

So plugging t in.. I get..
[tex]100000 sin \alpha cos \alpha = 15000 - 1587.302 sin \alpha[/tex]

How do I solve this for [tex]\alpha[/tex] ?
 
Last edited:
Use the 2 graph method, I just got this problem for homework myself. The angle should come out to around 8.6, and the time to around 21.4 seconds. Hope this helps
 

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