2 5 Motion Under Gravity – Class 11TH Physics

2.5 Motion Under Gravity

A body falling freely under the action of gravity is the most familiar example of uniformly accelerated rectilinear motion. According to Galileo, all bodies fall freely (in vacuum) under the acceleration due to gravity, denoted by ' $g$ '. Its experimental value is $9.8\,\mathrm{m/s^{2}}$ in SI units. This means that different bodies, when allowed to fall from the same height, strike the ground with the same velocity. As regards the sign of $g$ , it is taken positive for a falling body (when initial velocity is zero) and negative for a body projected vertically upward (when initial velocity is not zero).

The equations of motion for a freely falling body, on putting $a = g$ , become

v_{r} = v_{i} + gt

S = h = v_{i}t + \tfrac{1}{2}gt^{2}

v_{r}^{2} - v_{i}^{2} = 2gh

Example 2.5

An iron ball of mass $1\,\mathrm{kg}$ is dropped from a tower. The ball reaches the ground in $3.34\,\mathrm{s}$ . Find:

(a) the velocity of the ball on striking the ground,

(b) the height of the tower.

Solution

Since the ball is falling under the action of gravity, we shall put $a = g$ in equations of motion.

Mass of the ball $m = 1\,\mathrm{kg}$

Time taken to reach ground $t = 3.34\,\mathrm{s}$

Initial velocity $v_{i} = 0$

Final velocity $v_{r} = ?$

Acceleration $a = g = 9.8\,\mathrm{m/s^{2}}$

(a) Using first equation of motion:

v_{r} = v_{i} + gt

v_{r} = 0 + (9.8\,\mathrm{m\,s^{-2}})(3.34\,\mathrm{s})

\mathbf{v_{r} = 32.7\,\mathrm{m\,s^{-1}}}

(b) Using third equation of motion:

v_{r}^{2} - v_{i}^{2} = 2gh

(32.7\,\mathrm{m\,s^{-1}})^{2} - (0)^{2} = 2 \times 9.8\,\mathrm{m\,s^{-2}} \times h

h = \frac{1069.29\,\mathrm{m^{2}s^{-2}}}{19.6\,\mathrm{m\,s^{-2}}}

\mathbf{h = 54.56\,\mathrm{m}}

2.5 Motion Under Gravity

Example 2.5

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