Mechanics Projectiles Trajectory Question Help Needed

In summary, the anti aircraft gun with initial velocity 400m/s and angle theta fires at a pilotless aircraft flying at a speed of 100m/s and a constant height of 3500m. The equation of trajectory for the shell is y = xtan(theta) -0.5 gx^2/ (V^2 cos^2 (theta)), and using the given values, it can be simplified to x^2tan^2(theta)- (32000)xtan(theta) + (x^2 + 1.12x10^8) = 0. By substituting the values for y, V, and using the identity cos^2(theta)=1/(1+tan^2(theta)), the
  • #1
ishterz
14
0

Homework Statement


An anti aircraft gun with initial velocity 400m/s at angel theta above the horizontal, and the shells may be assumed to move freely under gravity. The target is a pilotless aircraft which flies at a speed of 100m/s directly towards the gun at a constant height of 3500m. A shell fired from the gun hits the aircraft when it is at a horizontal distance of 'x' m from the gun.

By using equation of trajectory show
x^2tan^2(theta)- (32000)xtan(theta) + (x^2 + 1.12x10^8) = 0


Homework Equations



Equation of trajectory: y = xtan(theta) - gx^2/ (V^2 cos^2 (theta))


The Attempt at a Solution



I assumed the y distance for both will be the same on collision, since the aircraft's height is constant

For the shell, I did :
x = 400cos(theta) t
therefore t= x/400cos(theta)

v= 400sin(theta)t - 5t^2

I subsituted for t in the second equation and tried to solve but could not get the answer.

Please help!

Thank you for your time
 
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  • #2
What is the question?

ehild
 
  • #3
How to show the above equation of x^2tan^2(theta)- (32000)xtan(theta) + (x^2 + 1.12x10^8) = 0
 
  • #4
What you did is the derivation of the equation of the trajectory, and your procedure is correct, if you meant y=400 sin(theta)-5t^2. But the equation you showed for the trajectory was not correct. It should be

y = xtan(theta) -0.5 gx^2/ (V^2 cos^2 (theta)).

You are also right using y=3500 m and V=400 m/s. Just plug in them to get the equation between theta and x, and use the identity

cos^2(theta)=1/(1+tan^2(theta)

to eliminate cos theta from the equation.

ehild
 
  • #5
Ahhh
Thank you!
 

FAQ: Mechanics Projectiles Trajectory Question Help Needed

1. What is a projectile in mechanics?

A projectile is any object that is launched into the air and moves along a curved path under the influence of gravity. Examples of projectiles include a ball thrown by a person, a bullet fired from a gun, or a rocket launched into space.

2. What is the trajectory of a projectile?

The trajectory of a projectile is the path it follows through the air. It is determined by the initial velocity, angle of launch, and the force of gravity. In the absence of air resistance, the trajectory of a projectile is a parabola.

3. How is the range of a projectile calculated?

The range of a projectile is the horizontal distance it travels before hitting the ground. It can be calculated using the formula R = (v^2 * sin(2θ))/g, where v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

4. How does air resistance affect the trajectory of a projectile?

Air resistance, also known as drag, can significantly affect the trajectory of a projectile. It acts in the opposite direction of the projectile's motion and can cause it to slow down and deviate from its expected path. As a result, the trajectory of a projectile with air resistance is not a perfect parabola.

5. What are some real-life applications of projectile motion?

Projectile motion is used in many real-life applications, such as the flight of a baseball, the trajectory of a golf ball, and the motion of a satellite in space. It is also essential in the design and functioning of weapons and sports equipment. Understanding projectile motion is crucial in fields such as engineering, physics, and sports science.

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