5.1 Basic terminologies for motion under gravity

  Motion in One Dimension

The basic terminologies for motion under gravity is,

1. Maximum height $\left( {{H_{\max }}} \right)$: The height up to which the body could reach is known as maximum height.

At maximum height, the velocity of a body becomes zero for a moment then its start increasing downwards.

Suppose a body is projected vertically upwards with velocity $u$.

Before proceeding we will define sign convention as,

So,

$\overrightarrow u = + u$ (Initial velocity is upwards)
$\overrightarrow a = - g$ (As $g$ acts in the downward direction)

At maximum height, velocity is zero.

$\overrightarrow v = 0$ (final velocity becomes zero at maximum height)

So, we can write the equation which relates velocity and displacement as, $${v^2} = {u^2} + 2\overrightarrow a .\overrightarrow s $$ or $${v^2} = {u^2} + 2\left( { - g} \right)\left( { + {H_{\max }}} \right)$$$${\overrightarrow H _{\max }} = + \frac{{{u^2}}}{{2g}}$$

Note: $+ve$ sign indicates that the displacement is in upward direction.

2. Time of ascent $\left( {{t_a}} \right):$ Time taken by body to reach the maximum height from the point of projection is known as time of ascent.

As we know at maximum height velocity is zero. So, we can write the situation as,

$\overrightarrow u = + u$
$\overrightarrow v = 0$
$\overrightarrow a = -g$

So, we can write the equation which relates velocity and time as,
$$\overrightarrow v = \overrightarrow u + \overrightarrow a t$$$$0 = + u - g{t_a}$$$${t_a} = \frac{u}{g}$$

3. Time of descent $\left( {{t_d}} \right):$ Time taken by body to reach the point of projection from the maximum height is known as time of descent.

As we know at maximum height velocity is zero. So, we can write the situation as,

$\overrightarrow u = + 0$
$\overrightarrow a = -g$
$\overrightarrow s = - {H_{\max }} = - \frac{{{u^2}}}{{2g}}$ (Body is going downward, so displacement is negative)

So, we can write the equation which relates displacement and time as, $$\overrightarrow s = \overrightarrow u t + \frac{1}{2}\overrightarrow a {t^2}$$$$ - \frac{{{u^2}}}{{2g}} = 0 - \frac{1}{2}g{\left( {{t_d}} \right)^2}$$$${\left( {{t_d}} \right)^2} = \frac{{{u^2}}}{{{g^2}}}$$$${t_d} = \frac{u}{g}$$

So, time of asent is equal to the time of descent. $${t_a} = {t_d} = \frac{u}{g}$$

4. Time of flight $(T)$: Time for which the particle is in air is known as time of flight.

In other words, the time for which the particle is in motion is also known as the time of flight.

So, time of flight is the sum of time of ascent and time of descent.
$$T = {t_a} + {t_d}$$$$T = \frac{u}{g} + \frac{u}{g}$$$$T = \frac{{2u}}{g}$$

5. Final velocity of the body after it comes back to the ground

$\overrightarrow u = + u$ (Initial velocity is upwards)
$t = \frac{{2u}}{g}$ (Time of flight)
$\overrightarrow a = - g$ (Acceleration due to gravity is downwards)

So, we can write the equation which relates velocity and time as,
$$\overrightarrow v = \overrightarrow u + \overrightarrow a t$$$$\overrightarrow v = + u + \left( { - g} \right)\frac{{2u}}{g}$$$$\overrightarrow v = - u$$
Note: $-ve$ sign indicates that the velocity is downwards,