Motion Of Body Under Gravity - Practice Questions & MCQ

The force of attraction of the earth on anybody is called the force of gravity. Acceleration produced on the body by the force of gravity is called acceleration due to gravity. It is represented by the symbol ‘g’.

Sign convention:

• Upward direction and right direction is taken as positive
• Downward direction and left direction is taken as negative

There are three cases basically in this - 

1) If a body dropped from some height (initial velocity zero)

$
\begin{aligned}
& \mathrm{u}=0 \\
& \mathrm{a}=\mathrm{g} \\
& v=g t \\
& h=\frac{1}{2} g t^2 \\
& v^2=2 g h \\
& h_n=\frac{g}{2}(2 n-1)
\end{aligned}
$

2) If a body is projected vertically downward with some initial velocity

$
\begin{aligned}
& \text { Equation of motion: } \quad v=u+g t \\
& \qquad \begin{array}{l}
h=u t+\frac{1}{2} g t^2 \\
v^2=u^2+2 g h \\
h_n=u+\frac{g}{2}(2 n-1)
\end{array}
\end{aligned}
$
 

3) If a body is projected vertically upward.
(i) Apply equation of motion :

Take initial position as origin and the direction of motion (vertically up) as $\mathrm{a}=-\mathrm{g} \quad$ [as acceleration due to gravity is downwards]
So, if the body is projected with velocity $u$, and after time t it reaches up to height h then,

$
\mathrm{v}=\mathrm{u}-\mathrm{g} t ; \quad \mathrm{h}=\mathrm{ut}-\frac{1}{2} \mathrm{gt}^2 ; \mathrm{v}^2=\mathrm{u}^2-2 \mathrm{gh}
$

(ii) For the case of maximum height $v=0$

So from the above equation

$
\begin{gathered}
u=g t \\
h=\frac{1}{2} g t^2 \\
\text { and } u^2=2 g h
\end{gathered}
$