###
__Projection & Motion Important
formulas:-__

__Projection & Motion Important formulas:-__

**Projection: Projectile fired at an angle with Horizontal:**

**It is the time taken by the projectile to return to ground, or the time for which the projectile remains in the air above the horizontal plane from the point of projection.**

__Time of flight:__
Time of ascent = u sinθ/g

Time of descent = u sinθ/g

And, Time of flight = Time of ascent +
Time of descent

Therefore, time of flight, T = 2u sin
θ/g

**It is the greatest height to which a projectile rises above the point of projection. it is represented by ‘H’.**

__Maximum Height Attend:__
Therefore, H = u² sin² θ/2g

**It is the distance covered by the projectile along the horizontal direction between the points of projection to the point on the ground where the projectile returns again. it is represented by ‘R’.**

__Horizontal Range:__
Therefore, R = u² sin2θ/g.

Where, u = Initial velocity in m/sec.

θ = angle of projection.

g = acceleration due to gravity
= 9.81 m/sec².

__Range and Time of flight on an inclined plane:__
Consider an inclined plane which makes
an angle θ

_{0}with the horizontal direction. Suppose an object projected with some initial velocity ‘u’ at an angle ‘θ’_{ }with the horizontal.
Let, the object strikes the inclined
plane at point ‘A’ after a time ‘T’.

Suppose, x-axis is taken along the
plane and y-axis is perpendicular to the plane.

The component of velocity along x-axis
and y-axis are;

u

_{x}= u cos(θ - θ_{0}) and u_{y}= u sin(θ_{ }- θ_{0})
The component of ‘g’ along x-axis and
y-axis are; (-g sinθ

_{0}) and (-g cosθ_{0})**T = {2u sin(θ**

__Time of flight:___{ }- θ

_{0})}/g. cosθ

**R = {2u² sin(θ**

__Range of projection on inclined plane:___{ }- θ

_{0}). cosθ}/g cos²θ

_{0}

__Notes:-__
1. For a given velocity of projection,
the horizontal range will be maximum, when

sin2θ = 1, or, 2θ = 90°,
or, θ = 45°

i.e. To achieve the maximum range, the
object should be projected at an angle of 45°

2. Horizontal range is same for angle
of projection θ and (90° - θ)

For angle of projection θ, horizontal range, R = u² sin2θ/g.

If the velocity of projection makes
angle θ with the vertical, then inclination with horizontal will be (90° - θ).

If R′ be the horizontal range for
angle of projection (90° - θ), then

R′ = u² sin 2(90° - θ)/g = u² sin (180
- 2θ)/g = u² sin2θ/g = R

Hence, the horizontal range is same
for an angle of projection θ and (90° - θ). Thus a football kicked at 30° or at 60° will
strike the ground at the same place, although when kicked at 60°, it will
remain longer in air.

3. If a body is projected from a place
above the surface of earth, then for the maximum range, the angle of projection
should be slightly less than 45°.

4. The trajectory of a projectile is a
parabola, only when the acceleration of the projectile is constant and the
direction of acceleration is different from the direction of velocity of the
projectile.

__Summary:__- 1. T = total time of flight = 2u sinθ/g
- 2. Time of ascent = Time of descent = u sinθ/g
- 3. Horizontal range = R = u² sin2θ/g
- 4. Horizontal coordinate, x = (u cosθ) × t or, R/2
- 5. Vertical coordinate, y = {(u sinθ) × t} -1/2gt² or, H = u² sin²θ/2g.

__Motion of a man in a lift:__**Case (1):**

__When the lift is accelerated upward__:

Suppose, ‘R’ be the upward thrust of
the floor on the man (normal reaction) and ‘mg’ is the weight of the man acting
downwards.

Hence, Unbalanced force = mass ×
acceleration

Or, R = mg + ma

Or, R = m (g + a)

Thus, if the man is standing on
weighting machine, it will show a larger weight than ‘mg’.

**Case (2):**

__When the lift is accelerate downwards:__

R+F = mg

Or, R
= mg - ma = m (g - a)

**Case (3):**

__When the lift moves with uniform velocity (or is at rest):__
In this case, a = 0 So, R = mg.

In the case of free fall of the lift,
a = g,

Then, r = m (g -
a), i.e. the man will feel weightlessness.

__Motion of a body vertically downward & vertically upward:__

__Motion of a body vertically downward:__

**from rest at a certain height h, then equation of motion are reduced to**

v = gt

h = ½gt²

v² = 2gh

Here, u = 0, s = h, a = +g.

[The equations of motion are: v = u +
at,

s = ut + ½at²,

v² = u² +2as.

Where, u = initial velocity

v = Final velocity

s = Distance Covered

a = Acceleration

g = Acceleration due to
gravity.]

If any of three quantities t, h and v
is given, then other two quantities can be determined.

__Motion of a body vertically upward:__

1. At time ‘t’, velocity of body is, v
= u - gt. [Here, a = -g]

2. At time ‘t’, the displacement of
body with respect to initial position is s = ut - ½gt²

3. The velocity of a body, when it has
a displacement ‘s’ is given by,

v² = u² - 2gs.

4. When it reaches maximum height from
‘A’, velocity, v = 0

Then, 0 = u - gt

Or, t = u/g at point B.

5. Maximum height attained by the
body, h = u²/2g

[Since, v² = u² - 2as, or, 0 = u² - 2gh, or, h = u²/2g]

6. Because displacement s = 0 at the
point of projection. Hence

s = ut - ½gt², or, 0 = ut - ½gt²,
or, t = 2u/g.

Therefore, Time of ascent = u/g

And, Time of descent = 2u/g - u/g =
u/g.

7. At any point ‘C’, between ‘A’ &
‘B’, where AC = s, the velocity ‘v’ is given by,

v = ± √(u² - 2gs)

This velocity of body whole crossing
point ‘C’, upward is +√(u² - 2gs) and while crossing ‘C’ downward is -√(u² -
2gs). The magnitude of velocity will remain same.

8. As, u = √2gh, hence time taken to
move up to highest point is also

u/g = √2gh/g =
√(2h/g).

__Motion on an inclined plane:__

- 1. Here, u = 0, a = g sinθ

Therefore, v = g sinθ × t
[Since v = u +at]

s = ½ (g sinθ × t²)
[Since, s= ut + ½at²]

v² = 2 g sinθ × s. [Since,
v² = u² +2as]

- 2. If ‘s’ is given, then, t² = 2s/(g sinθ).

**In the first ½ time, the body moves ¼th of the total distance, which in next half, it moves ¾th of the total distance on an inclined plane.**

__Note:__- 3. Time taken to move down on inclined plane:

**s = ½ g sinθ t² or, t = √(2s/g sinθ)**

As, h/s = sinθ or, s = h/sinθ

Hence, t = 1/sinθ × √(2h/g)

- 4. Because, v² = 2 g sinθ.s and, s = h/sinθ

Hence, v² = 2g sinθ × h/sinθ = 2gh

Or, v =√(2gh).

- 5. If friction is also present, but motion is taking along the inclined plane, then

F = ma = mg sinθ - µR

Or, F = mg sinθ - µmg cosθ.

Or, ma = m (g sinθ - µg cosθ) or, a = g (sinθ - µ cosθ) = g’

Therefore, v = √(2g’h)

And, t = 1/sinθ × √(2h/g’).

## 0 comments:

## Post a Comment