Projectile

A projectile is any object having a given initial velocity and moving along a path determined by the force of gravity and frictional resistance of the air. In an idealized situation, the object is assumed so small that it may be regarded as a particle. The air resistance is neglected. And the particle is supposed to be closed to the surface of the earth so that the acceleration due to gravity is almost constant.

Projectile Fired By Making An Angle With Horizontal

​Consider a projectile is fired with velocity $u$ from a point $O$ (origin) by making an angle $θ$ with the horizontal. $OX$ is X-axis and $OY$ is Y-axis. Then, the velocity $u$ can be resolved into two components; $u\cos θ$ along X-axis and $u\sin θ$ along Y-axis.

Projectile fired by making an angle with horizontal

Here,

  • $u\cos θ$ remains constant throughout the motion of the projectile.
  • $u\sin θ$ continuosly varies due to acceleration due to gravity.

And, the actual path followed by the projectile is known as trajectory.

The path followed by the projectile is parabolic.

​The motion of the projectile along Y-axis is, \[y=u_y t -\frac{1}{2}gt^2\] \[y=u\sin θ\cdot t -\frac{1}{2}gt^2\]

And, the motion of the projectile along X-axis is, \[x=u\cos θ ×t\] \[∴t=\frac{x}{u\cos θ}\]

Thus, we have, \[y=u\sin θ\cdot \frac{x}{u\cos θ}-\frac{1}{2}g\left(\frac{x}{u\cos θ}\right)^2\] \[y=\tan θ\cdot x-\left(\frac{g}{2u^2\cos^2θ}\right)x^2\]

This equation can be written as, \[y=ax-bx^2\]

This is a parabolic equation. Thus, the path followed by the projectile is parabolic. [Parabola]

Velocity of a projectile at a point

Let the mass of the projected particle be $m$ and suppose it is at the point $P(x, y)$ at the end of time $t$.

Let $F_x$ be the component of the force of gravity on the particle along X-axis and $a_x$ be the corresponding component of the acceleration due to gravity, then, by Newton’s Second Law of Motion, \[a_x=\frac{F_x}{m}=0\]

Similarly, if $F_y$ is the component of the force of gravity on the particle along Y-axis and $a_y$ be the corresponding component of the acceleration due to gravity, then, \[a_y=\frac{F_y}{m}=\frac{-mg}{m}=-g\]

Let $v$ be the velocity of the particle at $P(x, y)$ and $α$ be its inclination to the horizontal. Then, \[a_x=\frac{v\cos α-u\cos θ}{t}=0\] \[∴v\cos α=u\cos θ\] and, \[a_y=\frac{v\sin α-u\sin θ}{t}=-g\] \[∴v\sin α=u\sin θ-gt\]

Velocity at any height

We have, horizontal component of velocity $v$ at point $P$, \[v\cos α=u\cos θ\] and, vertical component of velocity $v$ at point $P$, \[v\sin α=u\sin θ-gt\]

Squaring and adding these two components, we get, \[v^2(\cos^2α+\sin^2α)=u^2(\cos^2θ+\sin^2θ)-2u\sin θgt+g^2t^2\] \[v^2=u^2-2u\sin θgt+g^2t^2\] \[v^2=u^2-2g(u\sin θt-\frac{1}{2}gt^2)\] \[∴v^2=u^2-2gy\]

Where, $y$ is the vertical height of the particle above $OX$. Also, \[\frac{v\sin α}{v\cos α}=\frac{u\sin θ-gt}{u\cos θ}\] \[\tan α=\frac{u\sin θ-gt}{u\cos θ}\] \[∴α=\tan^{-1}\left(\frac{u\sin θ-gt}{u\cos θ}\right)\]

Thus, from these equations, we can find both the magnitude and direction of velocity of the projectile at a given point or time.

Time of Flight

Time of flight is the total time taken by the projectile from its point of projection to its point of hitting the ground.

We have, \[y=u_yt-\frac{1}{2}gt^2\] \[y=u\sin θ\cdot t-\frac{1}{2}gt^2\]

When $t=T$ (Time of flight), then $y = 0$, \[0=u\sin θ\cdot t-\frac{1}{2}gt^2\] \[\frac{1}{2}gT^2=u\sin θT\] \[T=\frac{2u\sin θ}{g}\]

This is the time of flight of the projectile.

Maximum Height Attained

Maximum Height Attained By A Projectile

​The projectile, during its motion, attains a maximum height.

We have, \[v_y^2=u_y^2-2gh\]

When $h=h_{\text{max}}$ (maximum height), then $v_y=0$, \[0=u_y^2-2gh_{\text{max}}\] \[2gh_{\text{max}}=u^2\sin^2θ\] \[h_{\text{max}}=\frac{u^2\sin^2θ}{2g}\]

This gives the maximum height attained by the projectile.

Horizontal Range

Horizontal Range of a Projectile

​The total distance covered by the projectile from its point of projection to its point of hitting the ground is known as horizontal range. 

The distance covered by the projectile along X-axis is, \[x=u_xt\] \[x=u\cos θ\cdot t\]

When $t=T$ (Time of flight), then $x=R$ (horizontal range), \[∴R=u\cosθ\cdot T\]

We have, \[T=\frac{2u\sin θ}{g}\] \[∴R=u\cos θ\cdot \frac{2u\sin θ}{g}\] \[R=\frac{2u^2\sin θ\cos θ}{g}\] \[R=\frac{u^2\sin 2θ}{g}\]

This gives the horizontal range of the projectile. When $θ=45°$, then the horizontal range is maximum. i.e. \[R=\frac{u^2\sin(2×45)}{g}\] \[R=\frac{u^2\sin 90}{g}\] \[R=\frac{u²}{g}\]

Projectile Fired Horizontally From The Top Of A Tower

​Let us consider a projectile fired from the top of a tower of height $h$ with initial velocity $u$. Then, the component of initial velocity along X-axis. \[u_x = u\] and, the component of initial velocity along Y-axis. \[u_y = 0\]

Then, the X-componet of initial velocity remains constant throughout the motion. While, the Y-component of initial velocity changes with respect to time due to acceleration due to gravity.

Projectile fired horizontally from the top of a tower

Path of the Projectile

Suppose at any time $t$, the projectile is at a point $P(x,y)$. Then, the distance $x$ covered by the projectile in time $t$ is, \[x=u_xt\] \[x=ut\text{ ____(1)}\]

And, the distance $y$ covered by the projectile in time $t$ is, \[y=u_yt+\frac{1}{2}gt^2\] \[y=0+\frac{1}{2}gt^2\] \[y=\frac{1}{2}gt^2\]

From $(1)$, we have, \[t=\frac{x}{u}\] \[∴y=\frac{1}{2}g\frac{x^2}{u^2}\] \[y=\left(\frac{g}{2u^2}\right)x^2\] \[y=bx^2\]

This equation is parabolic. So, the path followed by the projectile is parabola.

Velocity of Projectile at P

The horizontal component of the velocity at time $t$ is given by, \[v_x=u_x=u\]

The vertical component of the velocity at time $t$ is given by, \[v_y=u_y+gt\] \[v_y=gt\] \[∴\text{Resultant Velocity (v)}=\sqrt{v_x^2+v_y^2}\] \[=\sqrt{u^2+g^2t^2}\]

And, the angle of resultant velocity, \[\tan θ=\frac{v_y}{v_x}\] \[\tan θ=\frac{gt}{u}\] \[θ=\tan^{-1}\left(\frac{gt}{u}\right)\]

Time of Flight

​The time taken by the projectile to reach to the ground is known as time of flight. We have, \[y=\frac{1}{2}gt^2\]

When $y=h$ (height of tower), then $t=T$ (time of flight) \[∴h=\frac{1}{2}gT^2\] \[∴T=\sqrt{\frac{2h}{g}}\]

Velocity of the projectile when it reaches the ground, We have, \[v=\sqrt{u^2+g^2t^2}\]

When the projectile reaches the ground, $t=T$, \[v=\sqrt{u^2+g^2T^2}\] \[v=\sqrt{u^2+g^2\frac{2h}{g}}\] \[v=\sqrt{u^2+2gh}\]

Also, direction of velocity at the ground, \[\tan θ=\frac{gT}{u}\] \[\tan θ=\frac{g\sqrt{\frac{2h}{g}}}{u}\] \[\tan θ=\frac{\sqrt{2gh}}{u}\] \[θ=\tan^{-1}\left(\frac{\sqrt{2gh}}{u}\right)\]

Horizontal Range

Horizontal Range $(R)$ is the horizontal distance covered by the projectile in the time of flight. We have, \[x=ut\] When $t=T$, $x=R$, \[R=uT\] \[R=u\sqrt{\frac{2h}{g}}\]


Q. For a given velocity of projection and a given horizontal range, prove that there are, in general, two directions of projection which are equally inclined to the direction of maximum range.

Two directions of projection equally inclined to the direction of maximum range

Let $u$ be the velocity of projection, $R$ be the horizontal range and $θ$ be the angle of projection. Then, \[R=\frac{u^2}{g}\sin 2θ\] \[R=\frac{u^2}{g}\sin(π-2θ)\] \[∴R=\frac{u^2}{g}\sin 2\left(\frac{π}{2}-θ\right)\] i.e. the value of the range remains same if $θ$ is replaced by $\frac{π}{2}-θ$.

Moreover, these two directions are equally inclined to the horizontal and vertical axes respectively. Hence, the direction of maximum range $(45°)$ bisects the angle between these two directions.

Q. A ball is thrown from the top of a building towards a tall building $50\sqrt{3}$ $\text{m}$ away. The initial velocity of the ball is $20$ $\text{m/s}$ at $30°$ above the horizontal. How far above or below its original level will the ball strike the opposite wall? (Take $g = 10$ $\text{m/s}^2$)

Question - Projectile

Let the ball thrown away from the top $A$ of a building strike another building at a point $B$ after $t$ seconds. Let $y$ be the distance of $B$ from the original level.

Then, horizontal component of initial velocity \[=20\cos 30=10\sqrt{3} \text{ ms}^{-1}\] And, vertical component of initial velocity \[=20\sin 30=10 \text{ ms}^{-1}\]

And, for the horizontal motion, \[50\sqrt{3}=10\sqrt{3}×t\] \[t=5\text{ secs}\]

During these $5$ seconds, the vertical distance covered, \[y=10.5+\frac{1}{2}(-10)×25=-75 \text{ m}\]

The negative sign shows that it strikes the opposite wall $75$ $\text{m}$ below the original level.


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