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A projectile is aimed at a mark on a horizontal plane through the point of projection and falls 6m short when elevation angle is 30°. The projectile overshoot the mark by 9m when angle of projection is 45°. The correct distance of the mark is ?
A projectile is aimed at a mark on a horizontal plane through the point of projection and falls 6m short when elevation angle is 30°. The projectile overshoot the mark by 9m when angle of projection is 45°. The correct distance of the mark is ?


4 years ago

Shaswata Biswas
132 Points
							Let the velocity of projection be u. Distance of mark from point of projection is x.When angle of projection is 30০ :Horizontal range = x - 6$\dpi{120} \implies \frac{u^{2}sin2*30^{o}}{g} = x - 6$           ....... (1)When the angle of projection is 45০ :Horizontal range = x + 9$\dpi{120} \implies \frac{u^{2}sin2*45^{o}}{g} = x + 9$            .........  (2)(1)÷(2)$\dpi{120} \frac{sin60^{o}}{sin90^{o}} = \frac{x -6}{x+9}$$\dpi{120} \implies x = \frac{9\sqrt{3}+12}{2-\sqrt{3}}$

4 years ago
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