Question

A particle is projected vertically upwards with velocity 40 m/s. Take $g=10m/s^2.$ Find the displacement and distance traveled by the particle during its time of flight.

Answer 1

Time of flight is $t=\frac{2u}{g}=\frac{2\times 40}{10}=8sec$

So time taken to reach at maximum height is $t_1=t/2=4sec$

Maximum vertical height $d=u{t}_1-\frac{g{t}_1^2}{2}$

$d=40\times 4-\frac{10\times 4^2}{2}=80m$

So after time 4 sec displacement is 80 m and also the distance covered is 80m.

After 8 sec it will reach back to ground, so displacement is zero and total distance traveled is $2d=2\times 80=160m$