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Question: An anti-tank gun fires a shell with a velocity of \(98\sqrt 2 m/s\) at an elevation of 45\(^\circ \)...

An anti-tank gun fires a shell with a velocity of 982m/s98\sqrt 2 m/s at an elevation of 45∘^\circ to the horizontal and the shell strikes a tank which is advancing with a velocity of 30 m/s directly towards the gun in the same horizontal plane as the gun. Find the distance between the gun and the tank at the instant when the gun is fired. g=9.8m/s2g = 9.8m/{s^2}

Explanation

Solution

The fired shell acts like a projectile fired at angle to the horizontal. We can calculate the maximum distance covered and the time of flight for the projectile. We know the velocity of a tank and time of flight is equal to the time taken by the tank to move from its position when the gun is fired. From this we can calculate the distance travelled by the tank while the shell reaches the tank. Adding this distance from the maximum distance travelled by shell gives us the required answer.
Formula used:
For a projectile fired at angle to the horizontal, we have the following relations:
Maximum horizontal range: R=u2sin⁡2θgR = \dfrac{{{u^2}\sin 2\theta }}{g}
Time of flight: T=2usin⁡θgT = \dfrac{{2u\sin \theta }}{g}

Complete answer:
We are given a gun which fires a shell at an angle to the horizontal and the shell acts like a projectile. The initial velocity and the angle of projection of the projectile are given as
u=982m/s θ=45∘ \begin{gathered} u = 98\sqrt 2 m/s \\\ \theta = 45^\circ \\\ \end{gathered}
Now using these values, we can calculate the time of flight and the maximum horizontal range for the projectile. This can be done in the following way.
Maximum horizontal range:
R=u2sin⁡2θg=(982)2sin⁡(2×45∘)9.8 =98×98×29.8=1960m \begin{gathered} R = \dfrac{{{u^2}\sin 2\theta }}{g} = \dfrac{{{{\left( {98\sqrt 2 } \right)}^2}\sin \left( {2 \times 45^\circ } \right)}}{{9.8}} \\\ = \dfrac{{98 \times 98 \times 2}}{{9.8}} = 1960m \\\ \end{gathered}
Time of flight:
T=2usin⁡θg=2×982×sin⁡45∘9.8 =20s \begin{gathered} T = \dfrac{{2u\sin \theta }}{g} = \dfrac{{2 \times 98\sqrt 2 \times \sin 45^\circ }}{{9.8}} \\\ = 20s \\\ \end{gathered}
We are given the velocity with which the tank travels. It is given as
v=30m/sv = 30m/s
When the gun is fired, it takes time equal to T to reach the tank. During this time, the tank travelled a distance which is equal to vT towards the gun from its initial position when the gun was fired. This distance is equal to 30×20=600m30 \times 20 = 600m.
Now we can calculate the initial distance between the gun and the tank at the instant when the gun is fired. It is equal to the sum of the maximum horizontal range and the distance travelled by tank during the time taken by the shell to reach the tank =1960−600=2560m = 1960 - 600 = 2560m. This is the required solution to the question.

Note:
We have added the two distances because the tank is advancing towards the gun. If the tank has been receding away from the gun then the initial distance of the tank would have been equal to the difference between the maximum horizontal range and the distance travelled by the tank during the time of flight of shell.