Question

A particle is projected vertically upwards with an initial velocity of `40 m//s.` Find the displacement and distance covered by the particle in `6 s.` Take `g= 10 m//s^2.`

Answer 1

Correct Answer - A
In 4s, it reaches upto the highest point and then
changes its direction of motion.
`s=ut+1/2 at^2=(40)(6)+1/2(-10)(6)^2`
`=60 m`
`d=|s|_(0-4)+|s|_(4-6)=|u^2/(2g)|+|1/2 g(t-t_0)^2|`
`=(40)^2/(2xx10)+1/2xx10xx(6-4)^2=100 m`