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# A particle is projected with a velocity u such that its range on a horizontal plane is twice the greatest height attained by it. The range of the projectile will be (g is the acceleration due to gravity).

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## Step 1: GIven dataA particle is projected with a velocity u such that its range on a horizontal plane is twice the greatest height attained by it. $R=2H$Step 2: Formula used$R=\frac{{u}^{2}\mathrm{sin}2\theta }{g}\left[whereR=range,u=initialvelocity,\theta =anglemadebythevelocitywithhorizontal,g=accelerationduetogravity\right]\phantom{\rule{0ex}{0ex}}H=\frac{{u}^{2}{\mathrm{sin}}^{2}\theta }{2g}\left[whereH=maximumheight\right]$Step 3: Calculating the range of the projectileWhen a particle is projected at an angle to the horizontal, the particle experiences both horizontal and vertical displacement. Because there will be no force along the horizontal axis, velocity along that axis will remain constant during its motion, however, there will be a force along the vertical axis, causing velocity along that axis to change. The range is the maximum horizontal displacement a bullet can have, while the maximum height is the most vertical displacement a projectile can have. According to the given data $R=2H$. By substituting the formulas we have we will get$\frac{{u}^{2}\mathrm{sin}2\theta }{g}=\frac{2{u}^{2}{\mathrm{sin}}^{2}\theta }{2g}\phantom{\rule{0ex}{0ex}}\frac{{u}^{2}2\mathrm{sin}\theta \mathrm{cos}\theta }{g}=\frac{2{u}^{2}{\mathrm{sin}}^{2}\theta }{2g}\left[As,\mathrm{sin}2\theta =2\mathrm{sin}\theta \mathrm{cos}\theta \right]\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =2$After getting the tangent value we can get sinusoidal and cosecant values because from the tangent value we have opposite side and adjacent side, we have to apply Pythagoras theorem to get the hypotenuse and from that, we can get sin and cos values. $\mathrm{sin}\theta =\frac{2}{\sqrt{5}},\mathrm{cos}\theta =\frac{1}{\sqrt{5}}$Substituting these values in the range formula$R=\frac{{u}^{2}2\mathrm{sin}\theta \mathrm{cos}\theta }{g}\phantom{\rule{0ex}{0ex}}R=\frac{{u}^{2}2\left(\frac{2}{\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}}\right)}{g}\phantom{\rule{0ex}{0ex}}R=\frac{4{u}^{2}}{5g}$Hence, the range of the projectile will be $R=\frac{4{u}^{2}}{5g}$.

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