Question

The maximum range of a gun horizontal terrain is \[10\,{\text{km}}\]. If \[g = 10\,{\text{m/}}{{\text{s}}^2}\], what must be the muzzle velocity of the shell?
A. \[400\,{\text{m/s}}\]
B. \[200\,{\text{m/s}}\]
C. \[300\,{\text{m/s}}\]
D. \[50\,{\text{m/s}}\]

Answer 1

Hint: Use the formula for the maximum horizontal range of a projectile. This formula gives the relation between the maximum horizontal range of the projectile, velocity of projection of the projectile and acceleration due to gravity. Convert the unit of the given maximum horizontal range of the gun to the SI system of units and substitute all values in this formula and calculate muzzle velocity of the gun.

Formula used:
The maximum horizontal range \[R\] of a projectile is given by
\[R = \dfrac{{{u^2}}}{g}\] …… (1)
Here, \[u\] is the velocity of the projection of the projectile and \[g\] is the acceleration due to gravity.

Complete step by step answer:
We have given that the maximum horizontal range of a gun horizontal terrain is \[10\,{\text{km}}\].
\[ \Rightarrow R = 10\,{\text{km}}\]
Convert the unit of maximum horizontal range of the gun to the SI system of units.
\[R = \left( {10\,{\text{km}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)\]
\[ \Rightarrow R = {10^4}\,{\text{m}}\]
Hence, the maximum horizontal range of the gun is \[{10^4}\,{\text{m}}\].
We have also given that the value of acceleration due to gravity is \[10\,{\text{m/}}{{\text{s}}^2}\].
\[ \Rightarrow g = 10\,{\text{m/}}{{\text{s}}^2}\]
We are asked to calculate the muzzle velocity which is the velocity of projection of the shell of the bullet.We can calculate this muzzle velocity of the shell using equation (1).Rearrange equation (1) for the velocity of projection of the shell.
\[{u^2} = Rg\]
\[ \Rightarrow u = \sqrt {Rg} \]
Substitute \[{10^4}\,{\text{m}}\] for \[R\] and \[10\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in the above equation.
\[ \Rightarrow u = \sqrt {\left( {{{10}^4}\,{\text{m}}} \right)\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)} \]
\[ \Rightarrow u = \sqrt {{{10}^5}} \]
\[ \Rightarrow u = {10^2}\sqrt {10} \]
\[ \Rightarrow u = {10^2}\left( {3.1} \right)\]
\[ \Rightarrow u = 310\,{\text{m/s}}\]
\[ \therefore u \approx 300\,{\text{m/s}}\]
Therefore, the muzzle velocity of the gun is \[300\,{\text{m/s}}\].

Hence, the correct option is C.

Note:The students should not forget to convert the unit of the maximum horizontal range of the gun to the SI system of units as all the values used in the formula are in the SI system of units. The students may also use the expression for horizontal range of a projectile and derive the expression for the maximum horizontal range of the projectile.