A ball trown up vertically returns to the thrower after 10 second. Calculate
(i) the velocity with which it was thrown up.
(ii) the maximum height it reachesand
(iii) its poisition after 7 second.

Explanation

(i) Time of ascent=time of descent`=(10)/(2)=5s`.
From `v=u+at`
`0=u-9.8xx5 , u=49 m//s`.
This is the velocity with which it was thrown up.
(i) Max.height `=u^(2)/2g=(49xx49)/(2xx9.8)=122.5m`
(iii) From `s=ut+(1)/(2)at^(2)`
`x=0+(1)/(2)xx9.8(2)^(2)=19.6m`,from the top.
position of the ball after `7 sec=122.5-19.6=102.9`m above the ground.