Question

A particle is projected vertically upwards with a velocity of 20 m/sec. Find the time at which the distance travelled is twice the displacement

• A. $2+\sqrt{\frac{4}{3}}sec$
• B. 1 sec
• C. $2+\sqrt{\frac{3}{4}}$
• D. 3 sec

Answer 1

The correct option is A. $2+\sqrt{\frac{4}{3}}sec$.

From the figure below, at point C, the distance travelled is twice the displacement.

For AB, using the third equation of motion we have,
$v^2-u^2=2\left(g\right)\left(H_{max}\right)$
$\Rightarrow 0-\left(20\right)\left(20\right)=-2\left(10\right).\left(\frac{3s}{2}\right)$
$\Rightarrow s=\frac{20\times 20}{10\times 3}=\frac{40}{3}$
Now, using the second equation of motion we have,

$s=ut-\frac{1}{2}g{t}^2$
$\Rightarrow \frac{40}{3}=20t-\frac{1}{2}\times 10t^2$

$3t^2-12t+8=0$
On solving $t=2+\sqrt{\frac{4}{3}}$.