Question

A cannon and a target are apart and located at the same level. How soon will the shell launched with the initial velocity $240\;{{\rm{m}} {\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}$ reach the target in the absence of air drag?

Answer 1

In the projectile motion of an object the time of flight provides information about how long an object stayed in air before he touched the ground and the horizontal distance covered in that complete journey will be the range of that projectile motion.

Complete step by step answer:
It is given that the cannon is falling apart $5.10\;{\rm{km}}$ from the starting point, so the value of the range for this projectile motion is: $R = 5.10\;{\rm{km}} = 5100\;{\rm{m}}$
Let the cannon is projected at an angle $\theta$ ,
It is given that the initial velocity of the cannon is: $u = 240\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}}$
The expression for the time of flight of the cannon is given as follows,
$\Rightarrow T = \dfrac{{2u\sin \theta }}{g}$
Here, $g$ is the acceleration due to the gravity.
It is known that the value of acceleration due to the gravity is $9.8\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}}$ .
From the above expression we will find the value of $\sin \theta$ .
$\Rightarrow \sin \theta = \dfrac{{gT}}{{2u}}......................\;{\rm{(1)}}$
We know that the range of an object is the product of the object's horizontal velocity and the time of flight of the projectile motion.
The horizontal component of the velocity of the cannon is : $u\cos \theta$
Now, we write the expression for the range of the cannon fired,
$\Rightarrow R = u\cos \theta \times {\rm T}\\\ \Rightarrow \cos \theta = \dfrac{R}{{uT}}......................\;{\rm{(2)}}$
Now, we take square of both the equation and add them,
$\Rightarrow {\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = {\left( {\dfrac{R}{{uT}}} \right)^2} + {\left( {\dfrac{{gT}}{{2u}}} \right)^2}$
It is known that the value of ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2}$ is equal to $1$ .
Substitute all the values in the above expression,
$\Rightarrow 1 = {\left( {\dfrac{{5100}}{{240 \times T}}} \right)^2} + {\left( {\dfrac{{9.8\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right. } {{{\rm{s}}^{\rm{2}}}}} \times T}}{{2 \times 240}}} \right)^2}$
On solving the above expression, we have
$\Rightarrow {T^4} - 2399{T^2} + 1083299 = 0$
Now we solve further the above equation,
$\Rightarrow {T^2} = \dfrac{{2399 \pm \sqrt {{{\left( { - 2399} \right)}^2} - 4 \times 1 \times 1083299} }}{2}\\\ \Rightarrow {T^2} = \dfrac{{2399 \pm 1192.47}}{2}$
Taking positive sign first,

$$\Rightarrow {T^2} = \dfrac{{2399 + 1192.47}}{2}\\\ \Rightarrow {T^2} = 1795.73$$

Take root both side,

$$\Rightarrow T = \sqrt {1795.73} \\\ \Rightarrow T = 42.37\;{\rm{s}}$$

Taking negative sign,

$$\Rightarrow {T^2} = \dfrac{{2399 - 1192.47}}{2}\\\ \Rightarrow {T^2} = 603.26$$

Take root both side,

$$\Rightarrow T = \sqrt {603.26} \\\ \Rightarrow T = 24.56\;{\rm{s}}$$

Therefore, the cannon will reach the same level at $42.37\;{\rm{s}}$ or $24.56\;{\rm{s}}$ after the launch.

Note: In this question,we should know that, in the projectile motion, the maximum range of the object is possible only when the object is projected at the angle of $45^\circ$.As this information is very helpful for us in solving various problems and also remember the time period expression at the angle $45^\circ$.