The velocity of a projectile when it is at the greatest height is \({\left(\sqrt{{{2}/{5}}}\right)}\) times its velocity when it is at half of its greatest height. Determine its angle of projection.

Answer: A

Suppose the particle is projected with velocity u at an angle theta with the horizontal. Horizontal component of its velocity at all height will be \({u}{\cos{\theta}}\).

At the greatest height, the vertical component of velocity is zero, so the resultant velocity is

\({v}_{{1}}={u}{\cos{\theta}}\)

At half the greatest height during upward motion,

\({y}={h}/{2},{a}_{{y}}=-{g},{u}_{{y}}={u}{\sin{\theta}}\)

Using \({{v}_{{{y}}}^{{2}}}-{{u}_{{{y}}}^{{2}}}={2}{a}_{{y}}{y}\)

we get, \({{v}_{{{y}}}^{{2}}}-{u}^{{2}}{{\sin}^{{2}}\theta}={2}{\left(-{g}\right)}\frac{{h}}{{2}}\)

or \({{v}_{{{y}}}^{{2}}}={u}^{{2}}{{\sin}^{{2}}\theta}-{g}\times\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{{2}{g}}}=\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{2}}{\left[\because{h}=\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{{2}{g}}}\right]}\)

or \({v}_{{y}}=\frac{{{u}{\sin{\theta}}}}{{\sqrt{{2}}}}\)

Hence, resultant velocity at half of the greatest height is

\({v}_{{2}}={\left(\sqrt{{{{v}_{{{y}}}^{{2}}}}}+{\left({{v}_{{{y}}}^{{2}}}\right)}\right)}\)

\(={\left(\sqrt{{{u}^{{2}}{{\cos}^{{2}}\theta}+{\left(\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{2}}\right)}}}\right)}\)

Given , \(\frac{{v}_{{1}}}{{v}_{{2}}}={\left(\sqrt{{\frac{{2}}{{5}}}}\right)}\)

\(\therefore\frac{{{{v}_{{{1}}}^{{2}}}}}{{{{v}_{{{2}}}^{{2}}}}}=\frac{{{u}^{{2}}{{\cos}^{{2}}\theta}}}{{{u}^{{2}}{{\cos}^{{2}}\theta}+{\left(\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{2}}\right)}}}=\frac{{2}}{{5}}\)

or \(\frac{{1}}{{{1}+\frac{{1}}{{2}}{{\tan}^{{2}}\theta}}}=\frac{{2}}{{5}}\)

or \({2}+{{\tan}^{{2}}\theta}={5}{\quad\text{or}\quad}{{\tan}^{{2}}\theta}={3}\)

or \({\tan{\theta}}={\left(\sqrt{{3}}\right)}\)

\(\therefore\theta={60}^{\circ}\) .

Suppose the particle is projected with velocity u at an angle theta with the horizontal. Horizontal component of its velocity at all height will be \({u}{\cos{\theta}}\).

At the greatest height, the vertical component of velocity is zero, so the resultant velocity is

\({v}_{{1}}={u}{\cos{\theta}}\)

At half the greatest height during upward motion,

\({y}={h}/{2},{a}_{{y}}=-{g},{u}_{{y}}={u}{\sin{\theta}}\)

Using \({{v}_{{{y}}}^{{2}}}-{{u}_{{{y}}}^{{2}}}={2}{a}_{{y}}{y}\)

we get, \({{v}_{{{y}}}^{{2}}}-{u}^{{2}}{{\sin}^{{2}}\theta}={2}{\left(-{g}\right)}\frac{{h}}{{2}}\)

or \({{v}_{{{y}}}^{{2}}}={u}^{{2}}{{\sin}^{{2}}\theta}-{g}\times\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{{2}{g}}}=\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{2}}{\left[\because{h}=\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{{2}{g}}}\right]}\)

or \({v}_{{y}}=\frac{{{u}{\sin{\theta}}}}{{\sqrt{{2}}}}\)

Hence, resultant velocity at half of the greatest height is

\({v}_{{2}}={\left(\sqrt{{{{v}_{{{y}}}^{{2}}}}}+{\left({{v}_{{{y}}}^{{2}}}\right)}\right)}\)

\(={\left(\sqrt{{{u}^{{2}}{{\cos}^{{2}}\theta}+{\left(\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{2}}\right)}}}\right)}\)

Given , \(\frac{{v}_{{1}}}{{v}_{{2}}}={\left(\sqrt{{\frac{{2}}{{5}}}}\right)}\)

\(\therefore\frac{{{{v}_{{{1}}}^{{2}}}}}{{{{v}_{{{2}}}^{{2}}}}}=\frac{{{u}^{{2}}{{\cos}^{{2}}\theta}}}{{{u}^{{2}}{{\cos}^{{2}}\theta}+{\left(\frac{{{u}^{{2}}{{\sin}^{{2}}\theta}}}{{2}}\right)}}}=\frac{{2}}{{5}}\)

or \(\frac{{1}}{{{1}+\frac{{1}}{{2}}{{\tan}^{{2}}\theta}}}=\frac{{2}}{{5}}\)

or \({2}+{{\tan}^{{2}}\theta}={5}{\quad\text{or}\quad}{{\tan}^{{2}}\theta}={3}\)

or \({\tan{\theta}}={\left(\sqrt{{3}}\right)}\)

\(\therefore\theta={60}^{\circ}\) .

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