An object is thrown vertically upwards with an initial velocity of 40 `"m.s"^(-1)`. (i) How long will the object move upwards? (ii) What will be the maximum height attained? (iii) How much time will it taken to reach the ground ? (iv) When will the object be at a height of 25 m from the ground ? (v) What will be its velocity after 2 s? [` g= 9.8 "m.s"^(-2)]`.

Explanation

(i) Let the time taken for upward motion be t s. At maximum height its velocity is zero. From equation
v = u - gt, we get,
`0 = 40 - 9.8 xxt " ""or"`, t = 4.1 s.
(ii) Let the maximum height attained be h.
From equation `v^(2) = u^(2) -2 `gh , we get,
`0 = (40)^(2) - 2xx9.8 h " ""or", h = 81.6`m.
Let the time taken to reach the ground be `t_(1)` starting from the time of projection. Considering both upward and downward motions of the body and using the equation h = ut`-(1)/(2) "gt"^(2)` , we get
`0 = 40 t_(1)-(1)/(2)xx9.8 xxt_(1)^(2)` [ total displacement is zero in this case ]
`:. " " t_(1)` = 8.2 s.
(iv) Let the time after which the body is at a height of 25 m be x.
Hence from h = `ut - (1)/(2)" gt"^(2) ` we get,
`25 = 40x -(1)/(2). 9.8 x^(2) " ""or", 49x^(2) - 400x+250 = 0`
or, `" " x =(400pmsqrt((400)^(2)-4.49.250))/(2.49)=(400pm10.sqrt1110)/(2.49)`
or,`" "`x = 0.682 s and x = 7.481s
Two values of x signify that the object will be at a height of 25 m twice during its flight , once (x = 0.682) s while moving upwards and the next (x = 7.81s) during its downward motion.
(v) Let the velocity acquired 2 s after the projection be v.
`:. " " v = 40 - 9.8 xx2 = 20.4 "m.s"^(-1)`.