**A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A,…**

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## Question “A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A,…”

enemy ships. One shell hits ship A, which is close by, and the

other hits ship B, which is farther away. The two shells are fired

at the same speed. Assume that air resistance is negligible and

that the magnitude of the acceleration due to gravity is g.

* What shape is the trajectory (graph of y vs. x) of the

shells?

a.straight line

b.parabola

c.hyperbola

d. The shape cannot be determined.

*For two shells fired at the same speed which statement about the

horizontal distance traveled is correct?

a. The shell fired at a larger angle with respect to the horizontal

lands farther away.

b. The shell fired at an angle closest to 45 degrees lands farther

away.

c. The shell fired at a smaller angle with respect to the

horizontal lands farther away.

d. The lighter shell lands farther away.

Now, consider for the remaining parts of the question below that

both shells are fired at an angle greater than 45 degrees with

respect to the horizontal. Remember that enemy ship A is closer

than enemy ship B.

* Which shell is fired at the larger angle?

a. A

b. B

c. Both shells are fired at the same angle.

* Which shell is launched with a greater vertical velocity,

vy?

a. A

b. B

c. Both shells are launched with the same vertical velocity.

*Which shell is launched with a greater horizontal velocity,

vx?

a A

b.B

c.Both shells are launched with the same horizontal velocity.

*Which shell reaches the greater maximum height?

a.A

b.B

c.Both shells reach the same maximum height.

*Which shell has the longest travel time (time elapsed between

being fired and hitting the enemy ship)?

a. A

b. B

c. Both shells have the same travel time.

## Answer

Body undergoing projectile movement is the solution to this problem.

First, use the projectile motion equations to determine the relationship between the coordinates and the position.

Projectile equations can be used to calculate the horizontal distance traveled maximum height of projectile and travel times.

Projectile motion is the process by which shells are fired from battleships. Parabola is the shape of projectile motion. The object is fired at an angle from the horizontal travels at the same horizontal velocity as air resistance. Vertically, gravity pulls it down with acceleration thanks to gravityg

The equation of trajectory is for a projectile.

{\vec r_f} = {\vec r_i} + {\vec v_i}t + \frac{1}{2}\vec g{t^2}Here, {\vec r_i}

The x \beginarrayc\\x = v_it\cos \theta \\\\y = v_it\sin \theta - \frac12gt^2\\\endarray

Here, v_i

Parabola’s general equation is:

y = ax^2Here, x

The hyperbola equation is the following:

\fracx^2a^2 - \fracy^2b^2 = 1Here, x

The equation for straight lines is:

y = (mx + cHere, x

The x{\rm{ and }}y \begin{array}{l}\\{v_{xi}} = {v_i}\cos \theta \\\\{v_{yi}} = {v_i}\sin \theta \\\end{array}

Here, {v_{xi}}

Time for projectile movement or flight is the time it takes to fly.

T = \frac{{2{v_i}\sin \theta }}{g}A projectile can only reach a maximum height of,

h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}At angle 90^\circ , the maximum value for sine function is 1.

**(a)**

The equation of trajectory is for a projectile.

{\vec r_f} = {\vec r_i} + {\vec v_i}t + \frac{1}{2}\vec g{t^2}Here, {\vec r_i}

Use the code MathTag11

\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}Substitute t = \frac{x}{{{v_i}\cos \theta }} \begin{array}{c}\\y = {v_i}\left( {\frac{x}{{{v_i}\cos \theta }}} \right)\sin \theta - \frac{1}{2}g{\left( {\frac{x}{{{v_i}\cos \theta }}} \right)^2}\\\\ = x\frac{{\sin \theta }}{{\cos \theta }} - \frac{1}{2}g\left( {\frac{{{x^2}}}{{v_i^2{{\cos }^2}\theta }}} \right)\\\end{array}

It is evident that the trajectory is not straight if you compare the equation to the equation of circle.

The equation of trajectory is for a projectile.

{\vec r_f} = {\vec r_i} + {\vec v_i}t + \frac{1}{2}\vec g{t^2}Here, {\vec r_i}

Use the code MathTag11

\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}Substitute t = \frac{x}{{{v_i}\cos \theta }} \begin{array}{c}\\y = {v_i}\left( {\frac{x}{{{v_i}\cos \theta }}} \right)\sin \theta - \frac{1}{2}g{\left( {\frac{x}{{{v_i}\cos \theta }}} \right)^2}\\\\ = x\frac{{\sin \theta }}{{\cos \theta }} - \frac{1}{2}g\left( {\frac{{{x^2}}}{{v_i^2{{\cos }^2}\theta }}} \right)\\\end{array}

It is evident that the equation is a parabola when compared to the equation.

The equation of trajectory is for a projectile.

{\vec r_f} = {\vec r_i} + {\vec v_i}t + \frac{1}{2}\vec g{t^2}Here, {\vec r_i}

Use the code MathTag11

\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}Substitute t = \frac{x}{{{v_i}\cos \theta }} \begin{array}{c}\\y = {v_i}\left( {\frac{x}{{{v_i}\cos \theta }}} \right)\sin \theta - \frac{1}{2}g{\left( {\frac{x}{{{v_i}\cos \theta }}} \right)^2}\\\\ = x\frac{{\sin \theta }}{{\cos \theta }} - \frac{1}{2}g\left( {\frac{{{x^2}}}{{v_i^2{{\cos }^2}\theta }}} \right)\\\end{array}

It is evident that the equation is not hyperbola when compared with the equation of parabola.

The equation of trajectory is for a projectile.

{\vec r_f} = {\vec r_i} + {\vec v_i}t + \frac{1}{2}\vec g{t^2}Here, {\vec r_i}

Use the code MathTag11

\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}The trajectory can thus be determined.

**(b)**

Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}At angle 90^\circ , the maximum value for sine function is 1.

\begin{array}{c}\\2\theta = 90^\circ \\\\\theta = 45^\circ \\\end{array}The shell was fired at angles greater than 45^\circ

Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}At angle 90^\circ , the maximum value for sine function is 1.

\begin{array}{c}\\2\theta = 90^\circ \\\\\theta = 45^\circ \\\end{array}The angle at which the shell was fired was closer to 45^\circ

Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}At angle 90^\circ , the maximum value for sine function is 1.

\begin{array}{c}\\2\theta = 90^\circ \\\\\theta = 45^\circ \\\end{array}The angle at which the shell was fired was smaller than 45^\circ

Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}Horizontal distance is not affected by the mass of the shells. Different mass shells, fired at the same angle and with the same speed, will reach the same distance. The lighter shell will not land further away, therefore.

**(c)**

Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}The enemy ship A is closer to enemy ship B. This means that the horizontal distance traveled by shell A is smaller. The angle at which the shell was fired was greater than 45^\circ

Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}The enemy ship A is closer to enemy ship B. This means that the horizontal distance traveled by shell A is smaller. The angle at which the shell was fired was greater than 45^\circ

Projectiles travel horizontally for a distance of,

R = \frac{{v_i^2\sin 2\theta }}{g}The enemy ship A is closer to enemy ship B. This means that the horizontal distance traveled by shell A is smaller. The angle at which the shell was fired was greater than 45^\circ

**(d)**

The y {v_{yi}} = {v_i}\sin \theta

Here, {v_{yi}}

The angle launched determines the vertical component of velocity. The sine increases for larger angles until it reaches 90^\circ .

The y {v_{yi}} = {v_i}\sin \theta

Here, {v_{yi}}

The angle launched determines the vertical component of velocity. The sine increases for larger angles until it reaches 90^\circ .

The y {v_{yi}} = {v_i}\sin \theta

Here, {v_{yi}}

The angle launched determines the vertical component of velocity. The sine increases for larger angles until it reaches 90^\circ .

**(e)**

The x {v_{xi}} = {v_i}\cos \theta

Here, {v_{xi}}

The angle launched determines the cosine of horizontal component of velocity. The cosine decreases with larger angles until it reaches minimum 90^\circ .

The x {v_{xi}} = {v_i}\cos \theta

Here, {v_{xi}}

The angle launched determines the cosine of horizontal component of velocity. The cosine increases with larger angles until it reaches maximum at 90^\circ .

The x {v_{xi}} = {v_i}\cos \theta

Here, {v_{xi}}

The angle launched determines the cosine of horizontal component of velocity. The cosine increases with larger angles until it reaches maximum at 90^\circ .

**(f)**

A projectile can only reach a maximum height of,

h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}Maximum height is determined by the square of the sine angle launched. Only if shells are fired at the same speed as g

A projectile can only reach a maximum height of,

h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}Maximum height is determined by the square of the angle launched. Only if shells are fired at the same speed as g

A projectile can only reach a maximum height of,

h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}Maximum height is determined by the square of the sine angle launched. Only if shells are fired at the same speed as g

**(g)**

Time of flight, or travel time for projectile movements is

T = \frac{{2{v_i}\sin \theta }}{g}The travel time of the shell is dependent on the angle of launch. Only if both shells are launched at the same speed and g

Time of flight, or travel time for projectile movements is

T = \frac{{2{v_i}\sin \theta }}{g}The travel time of the shell is dependent on the angle of launch. Only if both shells are launched at the same speed and g

Time of flight, or travel time for projectile movements is

T = \frac{{2{v_i}\sin \theta }}{g}The travel time of the shell is dependent on the angle launched. Only if both shells are launched at the same speed and gAns Part a

Parabola is the shape of the trajectory.

## Conclusion

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