Question

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Answer 1

Angle of Elevation - Projectile Motion

• Let me write the question again to make it better understandable.

Question:-

• If the range of a gun, which fires a shell with muzzle speed v, is R, then the angle of elevation of the gun is -

Answer:-

• The shell is fired at an angle. So, the shell is a Projectile and its motion can be described as Projectile Motion.

• Here, the velocity of projection is v and the Range is R. Suppose the angle of projection, which would be the angle of elevation of gun, is $\theta$.

$\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\vector(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){ \theta }\put(0.5,1.8){v}\put(4.3,0.7){\vector(-1,0){4}}\put(4.3,0.7){\vector(1,0){2}}\put(3.3,0.3){R}\end{picture}}$

• We can easily find out the answer by using the formula of Range.

$\displaystyle R = \frac{v^2\sin 2\theta}{g} \\\\\\ \implies \sin 2\theta = \frac{gR}{v^2} \\\\\\ \implies 2\theta = \sin^{-1}\left(\frac{gR}{v^2}\right) \\\\\\ \implies \Large \boxed{\theta = \frac{1}{2}\sin^{-1}\left(\frac{gR}{v^2}\right)}$

• Thus, we found out the Angle of Elevation of the Gun.

$\rule{300}{1}$

Extra Info:—

• Derivation of Formula of Range

• Let's assume the Time of Flight to be T.

• We can break down the motion into X and Y components. The only acceleration acting is the gravitational acceleration in the negative Y direction.

• So, we can break down the x and y components of velocity as:

$v_x = v\cos\theta \\\\ v_y = v\sin \theta$

And, the positions along x and y direction will be:

$x = v_x t = v\cos\theta\ t \\\\ y = v_yt + \frac{1}{2}(-g)t^2 = v\sin\theta\ t - \frac{1}{2}gt^2$

• The Flight ends when the shell touches the ground again (y=0). We can find the Time of Flight by equating y = 0 and solving the quadratic equation.

$\displaystyle y = v\sin\theta\ t -\frac{1}{2}gt^2 \\\\\\ \implies 0 = t\left(v\sin\theta-\frac{1}{2}gt\right) \\\\\\ \implies t=0 \quad \text{OR} \quad t = \frac{2v\sin\theta}{g}$

• The t = 0 corresponds to the time of Projection. The other value is the Time of Flight, when the flight ends.

• $\implies T = \dfrac{2v\sin\theta}{g}$

• When the shell touches the ground again, its x-coordinate will represent Range. Thus, by putting t = T in the equation for x, we can find Range.

$\displaystyle x = v\cos\theta\ t \\\\\\ \implies R = v\cos\theta\ T \\\\\\ \implies R = v\cos\theta \frac{2v\sin\theta}{g} \\\\\\ \implies R = \frac{v^2 (2\sin\theta\cos\theta)}{g} \\\\\\ \implies \large\boxed{R = \frac{v^2\sin 2\theta}{g}}$

And we used this formula above to get the answer.

Answer 2

Answer:

Angle of Elevation - Projectile Motion

Let me write the question again to make it better understandable.

Question:-

If the range of a gun, which fires a shell with muzzle speed v, is R, then the angle of elevation of the gun is -

Answer:-

The shell is fired at an angle. So, the shell is a Projectile and its motion can be described as Projectile Motion.

Here, the velocity of projection is v and the Range is R. Suppose the angle of projection, which would be the angle of elevation of gun, is \thetaθ .

Explanation:

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$Angle of Elevation - Projectile Motion</p><p></p><p>Let me write the question again to make it better understandable.</p><p></p><p>Question:-</p><p></p><p>If the range of a gun, which fires a shell with muzzle speed v, is R, then the angle of elevation of the gun is -</p><p></p><p>Answer:-</p><p></p><p>The shell is fired at an angle. So, the shell is a Projectile and its motion can be described as Projectile Motion.</p><p></p><p>Here, the velocity of projection is v and the Range is R. Suppose the angle of projection, which would be the angle of elevation of gun, is \thetaθ .</p><p></p><p>$