Question

A battleship simultaneously fires two shells toward two identical 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 1

Concepts and reason

The concept used to solve the problem is body undergoing projectile motion.

First, find the relation between the position coordinates by using the projectile motion equations.

The horizontal distance travelled maximum height of the projectile and travel time can be computed using projectile equations.

Fundamentals

The shells fired from a battleship undergo projectile motion. The shape of trajectory for a projectile motion is always parabola as the object fired at some angle from the horizontal travels with same horizontal velocity as air resistance is negligible and vertically the gravity pulls it down with acceleration due to gravity$g$.

The equation of trajectory for a projectile is,

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

Here, ${\vec r_i}$ is the initial position vector, ${\vec v_i}$ is the initial velocity vector, $\vec g$ is the acceleration due to gravity, and $t$ is any arbitrary time.

The $x$component and $y$component of the position vector of the projectile are,

$\begin{array}{c}\\x = {v_i}t\cos \theta \\\\y = {v_i}t\sin \theta - \frac{1}{2}g{t^2}\\\end{array}$

Here, ${v_i}$ is the initial velocity, $g$ is the acceleration due to gravity, $t$ is any arbitrary time, and $\theta$ is the angle which initial velocity make with respect to horizontal.

The general equation of parabola is,

$y = a{x^2}$

Here, $x$ and $y$ are the coordinates, and $a{\rm{ and }}b$ are any constants such that $a > 0$.

The general equation of a hyperbola is,

$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$

Here, $x$ and $y$ are the coordinates, and $a{\rm{ and }}b$ are any constants such that $a > 0,b > 0{\rm{ and }}a \ne b$.

The general equation of straight line is,

$y = mx + c$

Here, $x$ and $y$ are the coordinates, $m$ is the slope, and $c$ is the intercept.

The $x{\rm{ and }}y$ components of the initial velocity is,

$\begin{array}{l}\\{v_{xi}} = {v_i}\cos \theta \\\\{v_{yi}} = {v_i}\sin \theta \\\end{array}$

Here, ${v_{xi}}$ is the $x$-component of the initial velocity, ${v_{yi}}$ is the $y$-component of the initial velocity, ${v_i}$ is the initial speed, and $\theta$ is the angle which initial velocity make with respect to horizontal.

The time of flight or time for which projectile moves is,

$T = \frac{{2{v_i}\sin \theta }}{g}$

The maximum height for a projectile is,

$h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}$

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The maximum value of sine function is 1 at an angle $90^\circ$. The value of cosine is more for small angles and sine is more for large angles.

(a)

The equation of trajectory for a projectile is,

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

Here, ${\vec r_i}$ is the initial position vector, ${\vec v_i}$ is the initial velocity vector, $\vec g$ is the acceleration due to gravity, and $t$ is any arbitrary time.

Use the $x$ component of the position coordinate and solve for time.

$\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}$

Substitute $t = \frac{x}{{{v_i}\cos \theta }}$ in the equation $y = {v_i}t\sin \theta - \frac{1}{2}g{t^2}$ and find the relation between $x$ and $y$ coordinates.

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

Compare the equation with equation of circle it is clear that the trajectory is not a straight line.

The equation of trajectory for a projectile is,

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

Here, ${\vec r_i}$ is the initial position vector, ${\vec v_i}$ is the initial velocity vector, $\vec g$ is the acceleration due to gravity, and $t$ is any arbitrary time.

Use the $x$ component of the position coordinate and solve for time.

$\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}$

Substitute $t = \frac{x}{{{v_i}\cos \theta }}$ in the equation $y = {v_i}t\sin \theta - \frac{1}{2}g{t^2}$ and find the relation between $x$ and $y$ coordinates.

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

Compare the equation with equation of parabola it is clear that the trajectory is a parabola.

The equation of trajectory for a projectile is,

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

Here, ${\vec r_i}$ is the initial position vector, ${\vec v_i}$ is the initial velocity vector, $\vec g$ is the acceleration due to gravity, and $t$ is any arbitrary time.

Use the $x$ component of the position coordinate and solve for time.

$\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}$

Substitute $t = \frac{x}{{{v_i}\cos \theta }}$ in the equation $y = {v_i}t\sin \theta - \frac{1}{2}g{t^2}$ and find the relation between $x$ and $y$ coordinates.

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

Compare the equation with equation of parabola it is clear that the trajectory is not a hyperbola.

The equation of trajectory for a projectile is,

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

Here, ${\vec r_i}$ is the initial position vector, ${\vec v_i}$ is the initial velocity vector, $\vec g$ is the acceleration due to gravity, and $t$ is any arbitrary time.

Use the $x$ component of the position coordinate and solve for time.

$\begin{array}{c}\\x = {v_i}t\cos \theta \\\\t = \frac{x}{{{v_i}\cos \theta }}\\\end{array}$

Substitute $t = \frac{x}{{{v_i}\cos \theta }}$ in the equation $y = {v_i}t\sin \theta - \frac{1}{2}g{t^2}$ and find the relation between $x$ and $y$ coordinates.

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

Thus, the trajectory can be determined.

(b)

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The maximum value of sine function is 1 at an angle $90^\circ$. So, the horizontal distance is maximum when the angle is $90^\circ$.

$\begin{array}{c}\\2\theta = 90^\circ \\\\\theta = 45^\circ \\\end{array}$

The shell fired at angles larger than $45^\circ$ will travel less distance as compared to the shells fired at $45^\circ$. As the value of sine function will decrease after and before $45^\circ$. The horizontal distance travelled is dependent on sine function only as shells are fired at same speed and $g$ is same. Therefore, shells fired at a larger angle with respect to the horizontal lands closer to the battleship and not farther away.

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The maximum value of sine function is 1 at an angle $90^\circ$. So, the horizontal distance is maximum when the angle is $90^\circ$.

$\begin{array}{c}\\2\theta = 90^\circ \\\\\theta = 45^\circ \\\end{array}$

The shell fired at angles closer to $45^\circ$ will travel more distance as compared to the shells fired at angles far more or less than $45^\circ$. As the value of sine function will decrease after and before $45^\circ$. The horizontal distance travelled is dependent on sine function only as shells are fired at same speed and $g$ is same. Therefore, shells fired at an angle closest to $45^\circ$ lands farther away.

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The maximum value of sine function is 1 at an angle $90^\circ$. So, the horizontal distance is maximum when the angle is $90^\circ$.

$\begin{array}{c}\\2\theta = 90^\circ \\\\\theta = 45^\circ \\\end{array}$

The shell fired at angles smaller than $45^\circ$ will travel less distance as compared to the shells fired at $45^\circ$. As the value of sine function will decrease after and before $45^\circ$. The horizontal distance travelled is dependent on sine function only as shells are fired at same speed and $g$ is same. Therefore, shells fired at a smaller angle with respect to the horizontal lands closer to the battleship and not farther away.

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The horizontal distance does not depend on the mass of the shells. The shells of different mass but with same speed and fired at same angle will land at same distance. Therefore, the lighter shell does not land farther away.

(c)

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The enemy ship A is closer than enemy ship B. This means the shell A has travelled smaller horizontal distance. The shell fired at angles larger than $45^\circ$ will travel less distance as compared to the shells fired at $45^\circ$. As the value of sine function will decrease as angles goes more and more large after $45^\circ$. The horizontal distance travelled is dependent on sine function only as shells are fired at same speed and $g$ is same. Therefore, shells fired at a larger angle with respect to the horizontal lands closer to the battleship. Hence, the shell A is fired at larger angle.

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The enemy ship A is closer than enemy ship B. This means the shell A has travelled smaller horizontal distance. The shell fired at angles larger than $45^\circ$ will travel less distance as compared to the shells fired at $45^\circ$. As the value of sine function will decrease as angles goes more and more large after $45^\circ$. The horizontal distance travelled is dependent on sine function only as shells are fired at same speed and $g$ is same. Therefore, shells fired at a larger angle with respect to the horizontal lands closer to the battleship. Thus, the shell B is not fired at larger angle.

The horizontal distance travelled by a projectile is,

$R = \frac{{v_i^2\sin 2\theta }}{g}$

The enemy ship A is closer than enemy ship B. This means the shell A has travelled smaller horizontal distance. The shell fired at angles larger than $45^\circ$ will travel less distance as compared to the shells fired at $45^\circ$. As the value of sine function will decrease as angles goes more and more large after $45^\circ$. The horizontal distance travelled is dependent on sine function only as shells are fired at same speed and $g$ is same. Therefore, shells fired at a larger angle with respect to the horizontal lands closer to the battleship. Thus, the shells are fired at different angle.

(d)

The $y$ or vertical component of the initial velocity is,

${v_{yi}} = {v_i}\sin \theta$

Here, ${v_{yi}}$ is the $y$-component of the initial velocity, ${v_i}$ is the initial speed, $\theta$ is the angle which initial velocity make with respect to horizontal.

The vertical component of the velocity depends on sine of the angle launched. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell A is launched at larger angle. Hence, the vertical velocity ${v_y}$ is greater of the A shell.

The $y$ or vertical component of the initial velocity is,

${v_{yi}} = {v_i}\sin \theta$

Here, ${v_{yi}}$ is the $y$-component of the initial velocity, ${v_i}$ is the initial speed, $\theta$ is the angle which initial velocity make with respect to horizontal.

The vertical component of the velocity depends on sine of the angle launched. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell B is launched at smaller angle. Hence, the vertical velocity ${v_y}$ is smaller of the B shell.

The $y$ or vertical component of the initial velocity is,

${v_{yi}} = {v_i}\sin \theta$

Here, ${v_{yi}}$ is the $y$-component of the initial velocity, ${v_i}$ is the initial speed, $\theta$ is the angle which initial velocity make with respect to horizontal.

The vertical component of the velocity depends on sine of the angle launched. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell A and B are launched at different angles. Hence, the vertical velocity ${v_y}$ is different for both shells.

(e)

The $x$ or horizontal component of the initial velocity is,

${v_{xi}} = {v_i}\cos \theta$

Here, ${v_{xi}}$ is the $x$-component of the initial velocity, ${v_i}$ is the initial speed, $\theta$ is the angle which initial velocity make with respect to horizontal.

The horizontal component of the velocity depends on cosine of the angle launched. The cosine is smaller for larger angles till it reaches minimum at $90^\circ$. The shell A is launched at larger angle. Hence, the horizontal velocity ${v_x}$ is smaller for the A shell.

The $x$ or horizontal component of the initial velocity is,

${v_{xi}} = {v_i}\cos \theta$

Here, ${v_{xi}}$ is the $x$-component of the initial velocity, ${v_i}$ is the initial speed, $\theta$ is the angle which initial velocity make with respect to horizontal.

The horizontal component of the velocity depends on cosine of the angle launched. The cosine is smaller for larger angles till it reaches maximum at $90^\circ$. The shell A is launched at larger angle. Hence, the horizontal velocity ${v_x}$ is greater for the B shell.

The $x$ or horizontal component of the initial velocity is,

${v_{xi}} = {v_i}\cos \theta$

Here, ${v_{xi}}$ is the $x$-component of the initial velocity, ${v_i}$ is the initial speed, $\theta$ is the angle which initial velocity make with respect to horizontal.

The horizontal component of the velocity depends on cosine of the angle launched. The cosine is smaller for larger angles till it reaches maximum at $90^\circ$. The shell A and B are launched at different angles. Hence, the horizontal velocity ${v_x}$ is different for both shells.

(f)

The maximum height for a projectile is,

$h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}$

The maximum height depends on square of the sine of the angle launched only as shells are fired at same speed and $g$ is same. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell A is launched at larger angle. Hence, the maximum height is greater of the A shell.

The maximum height for a projectile is,

$h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}$

The maximum height depends on square of the sine of the angle launched only as shells are fired at same speed and $g$ is same. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell B is launched at smaller angle. Hence, the maximum height is smaller of the B shell.

The maximum height for a projectile is,

$h = \frac{{v_i^2{{\sin }^2}\theta }}{{2g}}$

The maximum height depends on square of the sine of the angle launched only as shells are fired at same speed and $g$ is same. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell A and B are launched at different angles. Hence, the maximum height is different for both shells.

(g)

The time of flight or travel time for which projectile moves is,

$T = \frac{{2{v_i}\sin \theta }}{g}$

The travel time of the shell depends on sine of the angle launched only as both shells are launched at same speed and $g$ is also same. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell A is launched at larger angle. Hence, the travel time is greater for the shell A.

The time of flight or travel time for which projectile moves is,

$T = \frac{{2{v_i}\sin \theta }}{g}$

The travel time of the shell depends on sine of the angle launched only as both shells are launched at same speed and $g$ is also same. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell B is launched at smaller angle. Hence, the travel time is smaller for the shell B.

The time of flight or travel time for which projectile moves is,

$T = \frac{{2{v_i}\sin \theta }}{g}$

The travel time of the shell depends on sine of the angle launched only as both shells are launched at same speed and $g$ is also same. The sine is more for large angles till it reaches maximum at $90^\circ$. The shell A and B are launched at different angles. Hence, the travel time is different for both shells.

Ans: Part a

The shape of the trajectory is parabola.