IEMDaily - Video Lecture Notes

Diagram

Time of Flight of a Projectile

Let, time taken to reach maximum height

= t_{m}

Now,

v_{x} = v_{o} \cos θ_{o}

and

v_{y} = v_{o} \sin θ_{o} - g t

Since, at this point,

v_{y} = 0

, we have:

v_{o} \sin θ_{o} - g t_{m} = 0

Or,

t_{m} = (v_{o} \sin θ_{o}) / g

Therefore, time of flight

= T_{f} = 2 t_{m} - 2 (v_{o} \sin θ_{o}) / g

because of symmetry of the parabolic path.

Diagram

Maximum Height of a Projectile

x = v_{x} t = (v_{o} \cos θ_{o}) t

and

y = (v_{o} \sin θ_{o}) t - (1 / 2) g t^{2}

The maximum height

h_{m}

is given by:

y = h_{m} = (v_{o} \sin θ_{o}) (\frac{v_{o} \sin θ_{o}}{g}) - \frac{g}{2} (\frac{v_{o} \sin θ_{o}}{g})^{2}

(for

t = t_{m}

)
Or,

h_{m} = \frac{(v_{o} \sin θ_{o})^{2}}{2 g}

Definition

Average Velocity of Projectile

The average velocity of a projectile between the instant it crosses one third the maximum height. It is projected with

u

making an angle

θ

with the vertical.
There will be a pair of points for which vertical velocities at the same height are in opposite direction and therefore their average sum

= 0

It is the horizontal velocity which is uniform and hence

v_{a v} = u_{x} = u \cos θ

For a general point:
Displacement in Y-direction:

y = u s i n θ \times t - \frac{g t^{2}}{2}

Displacement in X-direction:

x = u c o s θ \times t

Now in order to calculate average velocity:
Average Velocity =

\frac{N e t D i s p l a c e m e n t}{T o t a l t i m e}

Definition

Range of a Projectile

Range is maximum for

θ = 45^{0}

Moreover, from the expression of range:

R = \frac{u^{2} s i n 2 θ}{g}

Mathematically, it can be said that range is maximum for

θ = 45^{0}

And its maximum value will be

\frac{u^{2}}{g}

Definition

Relative Position of a Projectile with respect to a vertical obstacle

Let the co-ordinate of the pick point of the vertical obstacle is (x, y).
And at any instant of time, position of the projectile is (

v c o s θ

), (

v s i n θ

),
Position relative to the point at the obstacle is: (

v c o s θ - x

), (

v s i n θ - y

)
This is relative position of a projectile with respect to a vertical obstacle.

Result

Angle of Projection of a Projectile

For a projectile the range and maximum height are equal. The angle of projection is :Solution:
In projectile thrown at angle

θ

Range

R

and maximum height

H

are given as :
Range

R = \frac{u^{2} (\sin 2 θ)}{g} = \frac{u^{2} 2 \sin θ \cos θ}{g}

Max Height

H = \frac{u^{2} \sin^{2} θ}{2 g}

Given

H = R

\frac{u^{2} (2 \sin θ \cos θ)}{g} = \frac{u^{2} \sin^{2} θ}{2 g}

4cosθ=sinθ

4 \cos θ = s i n θ

\tan θ = 4

θ = 76^{0}

Definition

Magnitude of velocity at any point in projectile motion

Magnitude of velocity at any general point is:

v^{2} = v_{x}^{2} + v_{y}^{2}

Diagram

Angle of Projection of a Projectile with horizontal

v_{x} = v c o s θ \Rightarrow c o n s t a n t

v_{y} = v s i n θ - g t

Clearly

v_{y}

is maximum at t=0

Hence velocity,

v_{n e t} = \sqrt{v {_{x}}^{2} + v {_{y}}^{2}}

The direction of velocity will be along the direction of motion and angle of velocity with the horizontal will be given by:

t a n θ = \frac{v_{y}}{v_{x}}

Example

Radius of curvature of projectile at a point

Example:
A body is thrown from the surface of the Earth at an angle

α

to the horizontal with the initial velocity

v_{0}

. Assume that the air drag was neglected.Find radius of curvature of projectile. Assume acceleration due to gravity

= g

.Solution:
Since, radius of curvature

= \frac{{(v e l o c i t y)}^{2}}{n o r m a l a c c e l e r a t i o n}

At peak point A,velocity

= v_{o} \cos α

and normal acceleration

= g

R_{A} = \frac{v_{o}^{2} \cos^{2} α}{g}

Example

Equation of trajectory

Equation of trajectory can be found for a body whose displacement components are given as a function of time by eliminating the variable of time.
Example:
A radius vector of a point