IEMDaily - Video Lecture Notes

Formula

Equation of Parabolic Trajectory

y = x t a n θ - \frac{g x^{2}}{2 u^{2} c o s^{2} θ}

Where u is initial velocity and

θ

is angle of projection.

Definition

Path length of a projectile

For a projectile projected with an initial velocity

u

and initial angle with the horizontal

θ

, length of path for the projectile is given by:

l = \frac{u^{2} \cos^{2} θ}{2 g} [2 \sec θ \tan θ + l n | \frac{1 + \sin θ}{1 - \sin θ} |]

Example

Problem based on projectile motion on

Problem:
A car is moving horizontally along a straight line with constant speed

30 m / s

. A projectile is to be fired from the moving car in such a way that it will return to the car after the car has moved through

60 \sqrt{3} m

. The speed and angle at which the projectile must be projected are respectively:
Solution:
Range of projectile is given as

60 \sqrt{3}

, using formula for range we get,
Range

= 60 \sqrt{3} = \frac{u^{2} \sin 2 θ}{g}

As projectile is moving with same horizontal velocity as that of car, hence we can say that

u \cos θ = 30 - (1)

Solving equation of range

\Rightarrow 60 \sqrt{3} = \frac{2 u^{2} \sin θ \cos θ}{g}

\Rightarrow 30 \sqrt{3} = \frac{(2 u \sin θ) \times 30}{g}

\Rightarrow 10 \sqrt{3} = 2 u \sin θ

\Rightarrow u \sin θ = 5 \sqrt{3} - (2)

Solve (1) & (2) we get

u = 20 \sqrt{3}; θ = 30^{0}

Example

Projectile Projected from certain height above the ground

Example: A stone is projected from the top of a tower with velocity

20 m / s

making an angle of elevation of

30^{o}

with the horizontal. If the total time of flight is

5 s

and

g = 10 m / s^{2}

, then find the maximum height attained by this projectile.

Solution:

S_{y} = u t - \frac{1}{2} g t^{2}

h = S_{y} = 20 \sin 30^{o} \times 5 - \frac{1}{2} \times 10 \times (5)^{2}

h = 50 - 125 = - 75 m

(minus sign just indicates that the displacement is in downward direction)

h_{m a x} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{(20)^{2} \sin^{2} 30^{0}}{2 \times 10} = 5 m

Hence, maximum height attained by projectile

= h + h_{m a x} = 75 + 5 = 80 m

Example

Projectile clearing multiple obstacles

Example:
If a projectile crosses two walls of equal height

h

symmetrically as shown in the figure. Find time of flight, height of each wall and maximum height of projectile.

(g = 10 m / s^{2})

Solution:
Since the projectile motion is symmetric,

∴

Total time taken is,

T = 2 + 6 = 8 s

Also,

T = \frac{2 v \sin θ}{g}

and

H = \frac{v^{2} \sin^{2} θ}{2 g}

∴ H = \frac{g}{8} \frac{4 \times v^{2} \sin^{2} θ}{g^{2}}

∴ H = \frac{g \times T^{2}}{8} = \frac{10 \times 8^{2}}{8} = 80

Also, for

t = 2 s

assuming height of wall is

h

h = u \sin θ t - \frac{1}{2} g t^{2}

∴ h = 2 u \sin θ - 20

.....(i)
and,

2 u \cos θ = s

.....(ii)
for

t = 6 s

h = + 6 u \sin θ - 180

.....(iii)
and,

6 u \cos θ = s + 120

....(iv)
Equating (i) and (iii),

2 u \sin θ - 20 - 6 u \sin θ + 180 = 0

u \sin θ = \frac{160}{4} = 40

Put in equation (iii)

h = 60 m

Example

Projectile on an inclined Plane

An inclined plane is making an angle

β

with horizontal. A projectile is projected from the bottom of the plane with a speed

u

at an angle

α

with horizontal then its range

R

when it strikes the inclined plane is:
Time of flight,

O = u \sin (α - β) t - \frac{1}{2} g \cos β t^{2}

∴ t = \frac{2 u s i n (α - β)}{g c o s β}

Range

O B = (u c o s α) t = \frac{2 u^{2} s i n (α - β) c o s α}{g c o s β}

∴ O A = O B / c o s β = \frac{2 u^{2} s i n (α - β) c o s α}{g c o s^{2} β}

Example

Relative motion between an object in linear motion and a projectile

There is a toy plane moving horizontally at a height of 19.6m at a velocity of 50m/s. An object is dropped from the plane then the distance traveled by the plane at the instance object hits the ground has to be determined:

h = 1 / 2 g t^{2}

19.6 = 1 / 2 \times 9.8 \times t^{2}

t = 2 s

Distance moved by the plane in this interval of time is

50 \times 2 = 100 m

Example

Relative Motion between freely falling body and a projectile

Due to failure of engine of a rocket it started falling freely after

2

seconds an astronaut takes a horizontal velocity

u = 5

m/s to come out of it. Astronaut lands at a horizontal distance of

50

m from the rocket. The height of jump of the astronaut from the rocket has to be determined:

d = u \times t

50 = 5 \times t

t = 10 s

Initial velocity of the rocket at the instant of jump =

10 \times 2

h = u t + 1 / 2 g t^{2}

h = 5 \times 10 + 1 / 2 \times 10 \times 10^{2}

h = 550 m

Example

Relative motion in projectiles

Example:
CE and DF are two walls of equal height (

20

meter) from which two particles A and B of same mass are projected as shown in the figure.A is projected horizontally towards left while B is projected at an angle