Mechanics Projectiles Trajectory Question Help Needed

[ishterz]

1. The problem statement, all variables and given/known data
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


2. Relevant equations

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


3. 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

[ehild]

What is the question?

ehild

[ishterz]

How to show the above equation of x^2tan^2(theta)- (32000)xtan(theta) + (x^2 + 1.12x10^8) = 0

[ehild]

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

[ishterz]

Ahhh
Thank you!