Question

A particle is thrown with velocity $v_0$ in vertically upward direction from the ground. If particle is at same height at time $t=4s$ and $t=8s$, then the maximum height attained by the particle is $(g=10m/s^2$.

• A. $80m$
• B. $180m$
• C. $320m$
• D. $120m$

Answer 1

Answer: (B)

$180m$

Solution

Let time required to reach at the,

max height be t then,

$t=\left[\frac{t_1+t_2}{2}\right]=\frac{4+8}{2}=6$

To final initial vel, applying $I^{st}$

$e{q}^n$ of motion.

V = u+at

$\Rightarrow 0=u-\left(10\right)6$

$60m/s=u$

To find height, applying $II{I}^{rd}$

$e{q}^n$ of motion.

$\Rightarrow v^2-u^2=2as$

$\Rightarrow (0{)}^2-(60{)}^2=2(-10)H$

$\Rightarrow H=\frac{(60{)}^2}{20}=120m$

Option - D is correct.