Question

A particle is projected vertically upwards with an initial velocity of $404m/s$. Find the displacement and distance covered by the particle in $6$s. Take $g=10m/s^2.$

Answer 1

We know time of flight for an object $T=\frac{2v_0}{g}$ where g=acceleration due to gravity =10 m/$s^2$ and

$v_0$= velocity of the object =40 m/s.

Thus putting the above mentioned values in the given equation we get

$6=\frac{2\times 40}{10}\times \sin \theta$

$\theta$=${sin}^{-1}\frac{60}{80}$

$\theta ={48.59}^0$

The horizontal displacement covered by the particle is given by

$R=\frac{v_0^2\sin 2\theta }{g}$

$=\frac{{40}^2\sin 97.18}{10}$

$R=158.74m$