A body is thrown vertically upwards. After attaining half of its maximum height its velocity becomes 14 `"m.s"^(-2)`. (i) How high will the body rise? (ii)What will be the velocity of the body 1 s and 3 s after its projection? (iii)What is the average velocity of the body in the first haIf second?

Explanation

(i) Let the velocity of projection be u.
Hence maximum height attained by the body ,
`h = (u^(2))/(2g)` [ using `v^(2) = u^(2) -2gh ]`
or, `" "u^(2)` = 2gh
For half the maximum height i.e., `(h)/(2)`, we get
`(14)^(2) = u^(2) -2g""(h)/(2)=2gh - (1)/(2)(2gh)= gh`
`:.` h = 20 [ where g = 9.8 `"m.s"^(-2)`]
`:.` The body will rise up to a height of 20 m.
(ii) Here the velocity of projection,
`u = sqrt(2gh) = sqrt(2xx9.8xx20)=19.8 "m.s"^(-1)`
`:.` Velocity of the body 1 s after projection is ,
`v_(1) = u-g.1 = 19.8 - 9.8 = 10 "m.s"^(-1)`.
Velocity 3 s after projection,
`v_(2) = u - g xx 3 = 19.8 - 9.8xx3 = -9.6 "m.s"^(-1)`
(Negative sign indicates downward motion of the body.)
(iii) Velocity after `(1)/(2)` s,
`v. = u - gxx(1)/(2) = 19.8 - 9.8 xx (1)/(2) = 14.9 "m.s"^(-1).`
Hence average velocity during the given period
`(u+v.)/(2) = (19.8 +14.9)/(2) = 17.35 "m.s"^(-1)`