- Note
- Things to remember
- Videos

Any object thrown into atmosphere so that it falls under the effect of gravity alone is called projectile. Its path is trajectory. Here air resistance is neglected and acceleration due to gravity is downward. A projectile has two velocities horizontal and vertical velocity.

Suppose an object is projected with initial velocity u at an angle 90 with the ground taken as X-axis. OY is a vertical line perpendicular to the ground. The velocity in horizontal direction is \(u\cos \theta \) and in vertically upward direction \(u\sin \theta \) as shown in the figure.

**Path of Projectile**

Let the object is at a point P in time t whose horizontal and vertical distances are x and y . As horizontal velocity is constant, horizontal distance x in time t is given as

\begin{align*} x &= u\cos \theta \times t + 0 \: \text {as} \: a = 0 \\ t &= \frac {x}{u\cos \theta} \\ \text {here vertical velocity is affected by gravity} \\ y &= u\sin \theta \times t - \frac 12 gt^2 \\ &= u\sin \theta \times \frac {x}{u\cos \theta} - \frac 12 \times \frac {gx^2}{u^2\cos ^2 \theta } \\ y &= x\tan \theta - \frac {gx^2}{u^2\cos ^2 \theta } \\ \end{align*}

This equation is similar to \( y = ax + bx^2 \). This is an equation of a parabola. Hence the path of a projectile is parabolic.

**Time of Flight, T**

Time for which projectile remains is space. After time of flight h = 0

\begin{align*} h &= u\sin \theta t - \frac 12 gt^2 \: h = 0 \\ 0 &= t\left ( u\sin \theta - \frac 12 gt \right ) \text {or,} \: t = 0 \\ u\sin \theta &= \frac12 gt \\ \text{or,} \: t &= \frac {2u \sin \theta }{g} \end{align*}

**Horizontal Range, R**

Distance covered by the projectile during its time of flight. Since there is no acceleration in horizontal direction,

\begin{align*} \text {Horizontal range} &= \text {horizontal velocity} \times \text {time of flight} \\ \text {or,} \: R &= u \cos \theta \times \frac {2u \sin \theta }{g} \\ \text {or,} \: R &= \frac {u^2 \sin \theta \cos \theta }{g} \\ \therefore R &= \frac {u^2 \sin 2\theta }{g} \\ \end{align*}

**Maximum Range**

For a given initial velocity u, horizontal distance depends on angle of projection. Since \(\sin \theta \) has maximum value of 1, the horizontal range will be maximum when \(\sin 2\theta = 1\).

\begin{align*} \therefore \sin 2\theta &= 1 = \sin 90^o \\ \text {or,} \: 2\theta &= 90^o \\ \text {or,} \: \theta &= 45^o \\ \end{align*}

So, to achieve maximum horizontal range, the object must be projected at an angle of 45^{o} with the ground.

**Maximum height (H)**

A maximum height velocity becomes 0. So,

\begin{align*} V_y^2 &= (u\sin \theta)^2 – 2gH \\ 0 &= u^2\sin ^\theta -2gH \\ H &= \frac {u^2\sin ^2 \theta }{2g} \\ \end{align*}

**Two angles of Projection for the same Horizontal**

Let (θ) be angle of projection.

\begin{align*} R_1 &= \frac {u^2\sin 2 \theta }{g} \\ \text {If the angle of projection is} (90 - \theta ), R_2 \: \text {is} \\ R_2 &= \frac {u^2 \sin 2(90 - \theta)}{g} =\frac {u^2 \sin (180 – 2\theta )}{g} = \frac {u^2 \sin 2\theta }{g} \\ R_1 = R_2 \\ \end{align*}

So, the ranges are same in two angles of projection \(\theta \: \text {and} \: 90^o - \theta \) but time of flight at these angles of projection will be different.

An object is projected horizontally from height h above the ground with initial velocity u. The projectile is under the action of gravity. So, the horizontal velocity is constant and vertical velocity goes on increasing. (initial velocity = 0)

**Path of Projectile**

Let the position co-ordinate of the projectile at P be (x,y) after time t of its projection.

\begin{align*} \text {For horizontal distance,} \\ x &= u \times t \\ \therefore t &= \frac xu \\ \text {For vertical distance,} \\ y &= 0 + \frac 12 gt^2 \\ y &= \frac 12 g \times \left (\frac xu \right )^2 \\ \text {From above two equations, we have} \\ y &= \left (\frac {g}{2u} \right ) x^2 \\ \end{align*}

**Time of Flight**

The vertical distance covered by the projectile is equal to the height of the projection above the ground. If T is the time of flight, then

\begin{align*} h &= 0 + \frac 12 gT^2 \\ \text {or,} \: T &= \sqrt {\frac {2h}{g}}\\ \end{align*}

**Horizontal Range**

The horizontal range is the distance covered in horizontal direction in the time of flight T.

\begin{align*} R &= u \times T \\ \therefore R &= u \times \sqrt {\frac {2h}{g}}\\ \end{align*}

**Velocity of the Projectile at any instant**

Although the projectile initially has horizontal velocity, it gains vertical velocity due to the gravity. If v_{x} and v_{y} are the components of velocity at P in horizontal and vertical direction respectively after time t of its flight, the resultant velocity is given as

\begin{align*} v &= \sqrt {v_x^2 + v_y^2} \\ \text {Since} v_x \text {is constant being no acceleration in horizontal direction} \\ v_y &= 0 + gt \\ \text {or,} v_y &= gt \\ \therefore v &= \sqrt {v_x^2 + gt}\\ \text {If} \alpha \text {is the angle made by resultant velocity v with the horizontal, then}\\ \tan \alpha &= \frac {v_y}{v_x} \\ \text {or,} \: \alpha &= \tan ^{-1} \frac {gt}{u} \end{align*}

Any object thrown into atmosphere so that it falls under the effect of gravity alone is called projectile.

To achieve maximum horizontal range, the object must be projected at an angle of 45^{o} with the ground.

The horizontal range is the distance covered in horizontal direction in the time of flight T.

.-
## You scored /0

## ASK ANY QUESTION ON Projectile

You must login to reply

Sep 18, 2017

0 Replies

Successfully Posted ...

## Please Wait...

## asdfi

In what velocity falling ball with in infulence of gravity and from foot another ball thrown with velocity of 78.9m/s^2 and the height of projectile is 156m?

Mar 12, 2017

0 Replies

Successfully Posted ...

## Please Wait...

## samikshya

How high must a gun be aimed to hit a target at a distance R with velocity of bullet u?

Jan 30, 2017

0 Replies

Successfully Posted ...

## Please Wait...

## Prabhat

why is the initial velocity of projectile made from height zero but initial velocity of projectile from ground not zero?

Jan 30, 2017

1 Replies

Successfully Posted ...

## Please Wait...