TOPIC-3
Motion under gravity
- A body thrown vertically upwards or vertically downwards or dropped from a height will move in a straight vertical line.
- If air resistance is ignored, the 'body will be subjected to acceleration due to gravitational force exerted by the earth, which is denoted by g. The value of g on the earth is 9.8\(m/{s^2}\) in the downward direction.
- For small heights, the value of g is constant, we can use equations of uniformly accelerated motion.
- We shall take upward direction as positive & down direction as negative, as our convention.
Motion of a particle projected vertically upward from the ground
- Consider a particle projected vertically upward from the ground with velocity u.
Taking upward direction positive
\(u = u\)
\(a = - g\)
\(\therefore \) At any time t, velocity \(v = u + at\) and displacement \(s = ut - \frac{1}{2}g{t^2}\)
- To find time of ascent (\({t_a}\)) , apply \(v = u + at\) between the point of projection and the highest point,
To find total time of flight (T). apply \(s = ut + \frac{1}{2}a{t^2}\) between the point of projection and the time instant when the particle is again at point of projection
- Time of descent (\({t_d}\) ) between the highest point and back to the point of projection is also \(\frac{u}{g}\)
\(\therefore {t_d} = \frac{u}{g}\)
- For maximum height attained ( \({h_{\max }}\) ) apply \({v^2} = {u^2} + 2as\) between the point of projection and the topmost point,
\(\begin{array}{l}v = 0\\u = u\\a = - g\end{array}\)
\(\therefore {h_{\max }} = \frac{{{u^2}}}{{2g}}\)
- The particle will return back to the point of projection with same speed as the speed of projection but in the opposite direction.
- Motion under gravity is symmetric
In the above case speeds\({u_1}\) and \({u_2}\) are equal,\({t_{BC}} = {t_{CB}}\)\({t_{AB}} = {t_{BA}}\)
Motion of a particle projected downwards from height h above surface of earth
Suppose a particle is projected downwards from height h above the surface of the earth with speed u. To find the time taken by it to strike the surface of the earth, taking upward direction as positive,
\(\begin{array}{l}u = - u\\a = - g\\s = - h\end{array}\)
\(s = ut + \frac{1}{2}a{t^2}\)
solve the quadratic and get the positive value of t.
Motion of a particle protected vertically upwards from height h above surface of earth
\( - h = ut - \frac{1}{2}g{t^2}\), solve the quadratic and get the positive value of t.
Motion of a particle dropped from a height h above surface of earth
Solve using \({v^2} = {u^2} + 2as\) and \(s = ut + \frac{1}{2}a{t^2}\), taking
u = 0,
Velocity with which it strikes the surface will be \(\sqrt {2gh} \) and the time it will take to strike the surface will be \(\sqrt {\frac{{2h}}{g}} \).
