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A particle is aimed at a mark on the horizontal plane through the point of projection. It falls 12 meters short when the angle of projection is 15 degrees while it overshoots the mark by 24 meters when the angle is 45 degrees. Find the angle of projection to hit the mark. A particle is aimed at a mark on the horizontal plane through the point of projection. It falls 12 meters short when the angle of projection is 15 degrees while it overshoots the mark by 24 meters when the angle is 45 degrees. Find the angle of projection to hit the mark.
R’ is the required range = [u²/g] sin 2θ When θ = 15° R(15) = [u²/g] sin 30 = R’ – 12 where R’ is the required range .-----1 When θ = 45° R(45) = [u²/g] sin 45 = R’ + 24 where R’ is the required range ---2. 2-1gives 0.207[u²/g] = 36 [u²/g] = 173.9130 Substituting in 1 [u²/g] sin 30 = R’ – 12 173.9130* sin 30 = 173.9130 sin 2θ - 12 sin 2θ = 0.569 θ = 17.34° or 72.76° For these two angles the projectile will reach the mark.
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