Motion Of Body Under Gravity MCQ - Practice Questions with Answers

The force of attraction of earth on any body is called 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’.

There are three cases basically in this - 

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

                      u = 0 

                      a = g

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

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

Equation of motion:

$\quad v=u+g t$

$
\begin{array}{ll}
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}
$
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 $\mathrm{a}=-\mathrm{g}$ [As acceleration is downwards while motion upwards] So, if the body is projected with velocity $u$, and after time t it reaches up to height h then,

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

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

So from the above equation

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