What angles can a tank aim at to hit a target on a mountain 10km away?

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SUMMARY

The discussion focuses on calculating the angles at which a tank can aim to hit a target located 10km away on a mountain with a height of 500m. The tank's initial velocity is 500m/s, and the projectile motion equations are applied, specifically V(t)=Vo+at, X(t)=Xo+Vot+½at², and V²-Vo²=2aΔx. The challenge arises in solving for the angle θ due to the presence of both tanθ and cosθ in the equations, leading to a need for a common expression to simplify the calculations.

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devanlevin
a tank shoots at a target on a mountain 10km away from it, the height of the mountain is 500m the tank shoots at a speed of 1800km/h, at which angles can the tank aim?

this is a question in kinematics and the only force to take into account is gravity,


using
V(t)=Vo+at
X(t)=Xo+Vot+½at²
V²-Vo²=2aΔx

Vx=const=cosθ*Vo
Vy(t)=sinθ*Vo+at

Vo=500m/s
Xf=10000m
Yf=500m

knowing that the target is at (10000,500) i use the equation for X,
X(tf)=500cosθ*tf=10000m
tf=20/cosθ

Y(tf)=500m=500sinθ*(20/cosθ)-4.9*(20/cosθ)²
=1000tanθ-1960/(cosθ)²


now from here i don't know what to do, one equation with both tanθ and cosθ, how do i bring them to some common expression?? i have had a very similar quesion before which i too could not get past this point.

since i do not have T, i thought of using the 3rrd equation, but i don't see how i can
 
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Instead of solving in terms of theta, why not try solving in terms of t first?
 
how would you do that
 

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