Question

A particle is projected vertically upwards from point $A$ on the ground. It takes time $t_1$, to reach point $B$, but it still continues to move up. It takes further time $t_2$ to reach the ground from point $B$ then the height of point $B$ from the ground is:

Answer 1

Step. 1 Given Data,

Total time of particle = $(t_1+t_2)$

Step 2. Formula used,

Equations of motion, $v=u+at$,

Displacement Equation,$S=ut+{\displaystyle \frac{1}{2}}a{t}^2$.

Particle is projected vertically upwards so $u$ is the initial velocity

$a$ is the acceleration

$t$ is the time taken

$v$ is the final velocity

$$\begin{gathered} {\mathrm{t}}_1\mathrm{upward}\mathrm{motion}\mathrm{time} \\ {\mathrm{t}}_2\mathrm{downward}\mathrm{motion}\mathrm{time} \end{gathered}$$

Step 3. Calculation of height of point B from the ground

Time for A to B is $t_1$ and time from B to ground is $t_2$.

So, the time period for B to C would be $t_2-t_1$

Now applying eqn of motion for point A to a point where height is maximum

$v=u+at$

$\mathrm{v}=0\mathrm{at}\mathrm{maximum}\mathrm{height}$

Acceleration is considered $-g$
$v=u+at$.
$\Rightarrow 2u=g(t_1+t_2)$,

$u={\displaystyle \frac{1}{2}}g(t_1+t_2)$.
Initial velocity $u$ in the displacement equation of motion,

The equation is,
$h=u{t}_1-{\displaystyle \frac{1}{2}}g{t_1}^2$,
$h$ is the required height of point $B$.
Since

$u={\displaystyle \frac{1}{2}}g(t_1+t_2)$ ,

We will get,

$h=({\displaystyle \frac{1}{2}}g(t_1+t_2))t_1-{\displaystyle \frac{1}{2}}g{t_1}^2$
$\Rightarrow h={\displaystyle \frac{1}{2}}g{t_1}^2+{\displaystyle \frac{g{t}_1{t}_2}{2}}-{\displaystyle \frac{1}{2}}g{t_1}^2$
Therefore $h={\displaystyle \frac{g{t}_1{t}_2}{2}}$.
The height of the point $B$ from the ground is $\frac{g{t}_1{t}_2}{2}$.