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

AI Thread Summary
The discussion revolves around a projectile motion problem where a gun fires at a moving ship, requiring the calculation of the elevation angle and time to impact. Key equations for horizontal and vertical motion are provided, emphasizing the need to separate components and account for the ship's movement. A suggestion is made to use the quadratic formula to find the time of flight by setting the vertical displacement to zero at impact. The final calculations yield an elevation angle of approximately 8.6 degrees and a time interval of about 21.4 seconds. This approach effectively combines kinematic equations with the dynamics of a moving target.
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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 = \frac{2v_{0} sin\alpha }{g}

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

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



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.
 
40 \frac{km}{h} * 1000 \frac{m}{km} * \frac{1}{3600} \frac{h}{s} = 11.1 \frac{m}{s}

x_{max} = 15,000 - 11.1 t

x(t) = 700 cos \alpha t

700 cos \alpha t = 15000 - 11.1 t

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

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)
 
0 = \frac{-1}{2} (9.80) t^2 + 700 sin \alpha t

By the quadratic equation,

t = 142.86 sin \alpha

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

How do I solve this for \alpha ?
 
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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