**Vertical motion under gravity:**

Consider a body P projected vertically upwards from the surface of the earth with an initial velocity, u . The body rises to a maximum height at B and then returns (motion BC).

Even through we have shown the motion of the body as ABC, the actual path is always along the same line AB. The picture has been slightly modified for clarity.

Let the displacement of the body (at P) at time t, measured from its initial position A, be denoted by h.

We can now apply the equations . The acceleration, a = – g

(Note the positive direction in the figure, any vector in the opposite direction is negative);

x = h and the initial velocity is u . After time t ,

$ \displaystyle (i) v = u – g t $

$ \displaystyle (ii) h = ut – \frac{1}{2}g t^2 $

$ \displaystyle (iii) v^2 = u^2 – 2 g s $

### Other interesting questions that may be posed are :

(a) What’s the maximum height to which it rises ?

(b) What’s the time of flight ?

Let us note that at the point of maximum elevation , B , v_{B} = 0 (it’s got to be zero, if it were not the body would have risen further)

v_{B} = u – g t_{AB} = 0 ;

t_{AB} = time taken for motion AB.

$ \displaystyle t_{AB} = \frac{u}{g} $

If the maximum elevation is H (at the point B of course)

v_{B}^{2} = u^{2} − 2gH

0 = u^{2} − 2 g H

$ \displaystyle H = \frac{u^2}{2 g} $

When the body reaches the ground again (at the point C), we can write,

$ \displaystyle h = u t_{AC} – \frac{1}{2}g {t_{AC}}^2 $

Where t_{AC} = time taken for motion AC

$ \displaystyle 0 = u t_{AC} – \frac{1}{2}g {t_{AC}}^2 $

$ \displaystyle t_{AC} = \frac{2 u}{g} = 2 t_{AB} $

The velocity at the point C,

v_{C} = u − gt_{AC}

$ \displaystyle = u – g \times \frac{2 u}{g} $

= − u ;

which is equal in magnitude to u (the velocity of projection) but opposite in direction.

In both the examples considered above, the acceleration is constant (or uniform). This may not always be true.

__Particle projected from the top of a cliff__

For the motion of a particle projected vertically up or down from some height (h) we have assign the directions of displacement vector, velocity vector and acceleration vector with reference to a system of coordinates.

Let us consider the following cases :

(i) When the particle is projected with an initial velocity v_{0} in the upward direction and its vertical distance from the point of projection O at a time t be h, then

$ \displaystyle h = v_0 t – \frac{1}{2}g t^2 $ ; (if it is above the point O)

$ \displaystyle – h = v_0 t – \frac{1}{2}g t^2 $ ; (if it is below the point O)

We obtain the value of time by putting the given values of h and v_{0}.

If the origin is shifted to the ground level, which is at a depth H below the point of projection,

then h and – h mentioned above are replaced by (H + h) and (H – h) respectively.

**Solved Example : **A stone is dropped from a balloon ascending with v_{0} = 2 m/sec, from a height h = 4.8 m. Find the time of flight of the stone (g = 10 m/sec).

**Solution:** $\large h = – v_0 t + \frac{1}{2} g t^2 $

$\large t^2 – \frac{2 v_0}{g} t – \frac{2 h}{g} = 0 $

$\large t = \frac{v_0}{g} (1 + \sqrt{\frac{2 g h}{v_0^2} + 1 }) $

Putting the values of v_{0} and h , we obtain,

$\large t = \frac{2}{10} (1 + \sqrt{\frac{2 \times 10 \times 4.8}{2^2} + 1 }) $

t = 1.2 sec

Solved Example: Referring to the previous Problem , if the stone reaches the ground after t = 2 second from same initial height of release, find the

(a) speed of the balloon at the time of releasing the stone,

(b) total distance covered by the stone till it reaches the ground level,

(c) the average speed and

(d) average velocity of the stone for the total time of its flight, one from the bottom and the other from the top.

### Particle Projected from Top & Bottom

We see that s_{1} is downward ,

$ \displaystyle s_1 = v_1 t + \frac{1}{2}g t^2 $ ……(i)

$ \displaystyle s_2 = v_2 t – \frac{1}{2}g t^2 $ ……..(ii)

(i) + (ii) gives

⇒ s_{1} + s_{2} = (v_{1} + v_{2})t

$ \displaystyle t = \frac{h}{v_1 + v_2} $ …….(iii)

v_{1} + v_{2} = Relative velocity and

h = initial relative distance of separation.

**Solved Example : **A stone is released form the top of a cliff. Another particle is simultaneously projected with v = 20 m/sec. from the bottom of the cliff. If they meet after 2 seconds, find the height of the cliff.

**Solution: **Let after a time t, they meet s_{1} downward

$ \displaystyle s_1 = \frac{1}{2}g t^2 $ …..(1)

Since , s_{2} is upward

$ \displaystyle s_2 = v_2 t – \frac{1}{2}g t^2 $

Putting v_{2} = v ,we obtain ,

$ \displaystyle s_2 = v t – \frac{1}{2}g t^2 $ …(2)

(1) + (2)

⇒ s_{1} + s_{2} = v t

⇒ h = v t.

Putting v = 20 m/sec and t = 2 sec.

We obtain h = (20) (2) = 40 m.

Exercise : A body is released from a height and falls freely towards the earth. Exactly 1 sec. later another body is released. What is the distance between the bodies 2 sec. after the released of the second body if , g = 9.8 m/s^{2} ?

Exercise : Two particles are simultaneously projected upwards from the top and bottom of a cliff of height h = 20 m. If the speed of one is double that of the other and they meet after a time t = 2 second, find their speed of projection.