Recent content by mickstar

According to wikipedia, Using the pythagorean identity for cosθ in terms of tanθ equals Subbing this into 0.9 = 1250*tan(θ) - 11.975/cos²θ gives 0.9 = 1250tanθ - 11.975/ (±1/√(1 + tan2θ))2 0.9 = 1250tanθ ± 11.975 * (1 + tan2θ)2 oops, looks like i wrote it wrong..., WOW! i this...

How did you get 1.5625 and 11.975?

Thanks but Using this. i get 8002 = 8002sin2θ + 8002cos2θ = 8002(sin2θ + cos2θ) which is 8002 = 8002(1) because sin2θ + cos2θ = 1 for any theta making this a useless equation Ok so, using trig, i get Ux = 800cosθ substituting this into Δx = Ux * t gives 1250 = 800cosθ * t t = 1250/800cosθ Now...

Yes i have a sufficient knowledge of trig Ultimately the answer needed is θ, and I've been trying to get it through solving time. Through the formula Δx = Ux * t i have solved t to being 1250/800cosθ When i substitute this into Δy = Uy * t + a/2 * t^2 where Δy is 0.9m (1.2 - 0.3), a...

Here is the problem : A Sniper is lying in a prone position such that the barrel of his rifle is parallel to and 30cm above the ground. The muzzle velocity of his rifle is 800m/s A target appears at a distance of 1250m with the centre of the target 1.2m above ground level. If the sniper...