A body thrown vertically upwards with a certain velocity returns to the ground. What is its displacement and distance travelled ? How are the two affected when velocity of projection is doubled ?

Explanation

As the body returns to ground, displacement `= 0`.
We can show that height to which the body goes before falling down `= (u^(2))/(2g)`
`:.` Distance travelled in round trip `= 2 xx (u^(2))/(2g) = (u^(2))/(g)`.
When velocity of projection `(u)` is doubled, distance travelled becomes `4` times. However, displacement continues to be zero.