Projection & Motion Important formulas:-
Projection: Projectile fired at an angle with Horizontal:
Time of flight: 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 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
Maximum Height Attend: It is the greatest height to which a projectile rises
above the point of projection. it is represented by ‘H’.
Therefore, H = u² sin² θ/2g
Horizontal Range: 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’.
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 θ0 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;
ux = u cos(θ
- θ0) and uy = u sin(θ - θ0)
The component of ‘g’ along x-axis and
y-axis are; (-g sinθ0) and (-g cosθ0)
Time of flight: T = {2u sin(θ - θ0)}/g. cosθ
Range of projection on inclined plane: R = {2u² sin(θ - θ0).
cosθ}/g cos²θ0
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:
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).
Note: 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.
- 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’).