Projection and Motion

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;

                        u_x = u cos(θ - θ₀) and u_y = 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:

 When a body is released 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:

 Suppose a body is projected vertically upward from a point ‘A’ with an initial velocity ‘u’.

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’).