Answer
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Hint We are given here with the kinetic energy ratio and the ratio of maximum height and we are asked to find out the ratio of their ranges. So we will find the ratio of their velocities and angles of the projectile and use the formula for range.
Formula used
\[{E_k} = \dfrac{1}{2}m{u^2}\]
Where, \[{E_k}\] is the kinetic energy of the projectile, $m$ is the mass of the projectile and $u$ is the initial velocity of the projectile.
\[H = \dfrac{{{u^2}si{n^2}\theta }}{{2g}}\]
Where, \[H\] is the maximum height of the projectile, $u$ is the initial velocity of the projectile, $\theta $ is the angle of the projectile with the horizontal and $g$ is the acceleration due to gravity.
\[R = \dfrac{{{u^2}sin2\theta }}{g}\]
Where, $R$ is the range of the projectile, $u$ is the initial velocity of the projectile, $\theta $ is the angle of the projectile with the horizontal and $g$ is the acceleration due to gravity.
Complete Step By Step Solution
We are given,
\[\dfrac{{Kinetic{\text{ }}Energy{\text{ }}of{\text{ }}the{\text{ }}first{\text{ }}projectile}}{{Kinetic{\text{ }}Energy{\text{ }}of{\text{ }}the{\text{ second }}projectile}} = \dfrac{{{E_{k1}}}}{{{E_{k2}}}} = \dfrac{4}{1}\]
Thus, putting in the formula for kinetic energy, we can say
\[\dfrac{{\dfrac{1}{2}m{u_1}^2}}{{\dfrac{1}{2}m{u_2}^2}} = \dfrac{4}{1}\]
Thus, after cancellation, we get
\[\dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{4}{1}\]
Thus, we get
\[\dfrac{{{u_1}}}{{{u_2}}} = \dfrac{2}{1} \Rightarrow {u_1}:{u_2} = 2:1\]
Now,
\[\dfrac{{Maximum{\text{ }}Height{\text{ }}Of{\text{ }}the{\text{ }}first{\text{ }}projectile}}{{Maximum{\text{ }}Height{\text{ }}Of{\text{ }}the{\text{ second }}projectile}} = \dfrac{{\dfrac{{{u_1}^2si{n^2}{\theta _1}}}{{2g}}}}{{\dfrac{{{u_2}^2si{n^2}{\theta _2}}}{{2g}}}} = \dfrac{4}{1}\]
After cancellation and Putting in \[\dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{4}{1}\] , we get
\[\dfrac{{si{n^2}{\theta _1}}}{{si{n^2}{\theta _2}}} = \dfrac{1}{1}\]
Thus, we can say
${\theta _1} = {\theta _2}$
Now,
$\dfrac{{Range{\text{ }}of{\text{ }}the{\text{ }}first{\text{ }}projectile\;}}{{Range{\text{ }}of{\text{ }}the{\text{ second }}projectile\;}} = \dfrac{{\dfrac{{{u_1}^2\sin 2{\theta _1}}}{g}}}{{\dfrac{{{u_2}^2\sin 2{\theta _2}}}{g}}}$
After cancellation and putting in $\dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{4}{1}$ and $\dfrac{{{\theta _1}}}{{{\theta _2}}} = \dfrac{1}{1}$, we get\[Range{\text{ }}of{\text{ }}the{\text{ }}first{\text{ }}projectile:{\text{ }}Range{\text{ }}of{\text{ }}the{\text{ }}second{\text{ }}projectile = 4:1\]
Hence, the correct option is (B).
Note We evaluated the value of \[\dfrac{{{u_1}}}{{{u_2}}}\] and $\dfrac{{{u_1}^2}}{{{u_2}^2}}$. This was for being more precise with the answer. Moreover, the value of \[\dfrac{{{u_1}}}{{{u_2}}}\] could$ \pm \dfrac{2}{1}$. But the value of velocity of a projectile cannot be negative. Thus, we took the value of \[\dfrac{{{u_1}}}{{{u_2}}}\] to be $\dfrac{2}{1}$.
Formula used
\[{E_k} = \dfrac{1}{2}m{u^2}\]
Where, \[{E_k}\] is the kinetic energy of the projectile, $m$ is the mass of the projectile and $u$ is the initial velocity of the projectile.
\[H = \dfrac{{{u^2}si{n^2}\theta }}{{2g}}\]
Where, \[H\] is the maximum height of the projectile, $u$ is the initial velocity of the projectile, $\theta $ is the angle of the projectile with the horizontal and $g$ is the acceleration due to gravity.
\[R = \dfrac{{{u^2}sin2\theta }}{g}\]
Where, $R$ is the range of the projectile, $u$ is the initial velocity of the projectile, $\theta $ is the angle of the projectile with the horizontal and $g$ is the acceleration due to gravity.
Complete Step By Step Solution
We are given,
\[\dfrac{{Kinetic{\text{ }}Energy{\text{ }}of{\text{ }}the{\text{ }}first{\text{ }}projectile}}{{Kinetic{\text{ }}Energy{\text{ }}of{\text{ }}the{\text{ second }}projectile}} = \dfrac{{{E_{k1}}}}{{{E_{k2}}}} = \dfrac{4}{1}\]
Thus, putting in the formula for kinetic energy, we can say
\[\dfrac{{\dfrac{1}{2}m{u_1}^2}}{{\dfrac{1}{2}m{u_2}^2}} = \dfrac{4}{1}\]
Thus, after cancellation, we get
\[\dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{4}{1}\]
Thus, we get
\[\dfrac{{{u_1}}}{{{u_2}}} = \dfrac{2}{1} \Rightarrow {u_1}:{u_2} = 2:1\]
Now,
\[\dfrac{{Maximum{\text{ }}Height{\text{ }}Of{\text{ }}the{\text{ }}first{\text{ }}projectile}}{{Maximum{\text{ }}Height{\text{ }}Of{\text{ }}the{\text{ second }}projectile}} = \dfrac{{\dfrac{{{u_1}^2si{n^2}{\theta _1}}}{{2g}}}}{{\dfrac{{{u_2}^2si{n^2}{\theta _2}}}{{2g}}}} = \dfrac{4}{1}\]
After cancellation and Putting in \[\dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{4}{1}\] , we get
\[\dfrac{{si{n^2}{\theta _1}}}{{si{n^2}{\theta _2}}} = \dfrac{1}{1}\]
Thus, we can say
${\theta _1} = {\theta _2}$
Now,
$\dfrac{{Range{\text{ }}of{\text{ }}the{\text{ }}first{\text{ }}projectile\;}}{{Range{\text{ }}of{\text{ }}the{\text{ second }}projectile\;}} = \dfrac{{\dfrac{{{u_1}^2\sin 2{\theta _1}}}{g}}}{{\dfrac{{{u_2}^2\sin 2{\theta _2}}}{g}}}$
After cancellation and putting in $\dfrac{{{u_1}^2}}{{{u_2}^2}} = \dfrac{4}{1}$ and $\dfrac{{{\theta _1}}}{{{\theta _2}}} = \dfrac{1}{1}$, we get\[Range{\text{ }}of{\text{ }}the{\text{ }}first{\text{ }}projectile:{\text{ }}Range{\text{ }}of{\text{ }}the{\text{ }}second{\text{ }}projectile = 4:1\]
Hence, the correct option is (B).
Note We evaluated the value of \[\dfrac{{{u_1}}}{{{u_2}}}\] and $\dfrac{{{u_1}^2}}{{{u_2}^2}}$. This was for being more precise with the answer. Moreover, the value of \[\dfrac{{{u_1}}}{{{u_2}}}\] could$ \pm \dfrac{2}{1}$. But the value of velocity of a projectile cannot be negative. Thus, we took the value of \[\dfrac{{{u_1}}}{{{u_2}}}\] to be $\dfrac{2}{1}$.
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