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# An artillery piece which consistently shoots its shells with the same muzzle speed has a maximum range R. To hit a target which is from the gun and on the same level, the elevation angle of the gun should be.

Last updated date: 20th Jun 2024
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Hint: When a particle is thrown at an elevated angle near the earth’s surface, it moves in a curved path under constant acceleration that is directed towards the center of the earth (assuming that the particle remains close to the earth's surface). The path made by such a particle is called a projectile and the motion is known as projectile motion. The air resistance in the motion of the body is neglected in projectile motion.

Now from the question
The elevation angle of the gun is the angle between the horizontal plane and the axial direction of the barrel of a gun, mortar or heavy artillery.
According to the given question,
Range of projectile= R
And $R = \dfrac{{{u^2}\sin 2\theta }}{g}$
The Range of projectile is maximum when it is projected at an angle of $45^\circ$ and is given by,
$R = \dfrac{{{u^2}}}{g}$
Where $u$ is the velocity of projection
${u^2} = Rg$ ………………………………….(1)
Now when $R = \dfrac{R}{2}$
$\dfrac{R}{2} = \dfrac{{{u^2}\sin 2\theta }}{g}$
Where $\theta$ is the angle of projection
$\dfrac{R}{2} = \dfrac{{Rg\sin 2\theta }}{g}$ [from (1) ${u^2} = Rg$]
$\sin 2\theta = \dfrac{1}{2}$
$\sin 2\theta = \sin 30^\circ$
$\theta = 15^\circ$
Therefore, the elevation angle of the gun should be at $\theta = 15^\circ$

A projectile is an object that gives an initial velocity, gravity acts on it. The range of projectiles horizontal is the distance along the horizontal plane. When the horizontal range is depending upon the initial velocity $v$ , the angle$\theta$ and the acceleration occurring due to the gravity.