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

In summary, the required elevation angle for the gun is 8.6 degrees. The time interval between firing and impact is 21.4 seconds.
  • #1
Jacobpm64
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0

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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  • #2
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.
 
  • #3
[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.
 
  • #4
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)
 
  • #5
[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:
  • #6
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
 

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

1. What is projectile motion?

Projectile motion refers to the motion of an object that is launched into the air and moves along a curved path under the influence of gravity and air resistance.

2. How is the trajectory of a projectile calculated?

The trajectory of a projectile can be calculated using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and the angle of launch.

3. What factors affect the distance and height of a projectile?

The distance and height of a projectile are affected by the initial velocity, angle of launch, and the amount of air resistance or drag acting on the object.

4. How does projectile motion relate to real-world situations?

Projectile motion has many real-world applications, such as in sports like baseball and football, as well as in areas like engineering and physics. Understanding the principles of projectile motion can help in designing and predicting the motion of objects in these situations.

5. What is the impact of angle of launch on projectile motion?

The angle of launch plays a significant role in determining the range of a projectile. For example, a projectile launched at a 45-degree angle will have the maximum range, while a projectile launched at a 90-degree angle will have the maximum height but zero range.

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