A ball is thrown vertically upwards with a velocity of `20 ms^(-1)` from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground ? Take `g = 10 ms^(-2)`.

Explanation

(a) Suppose height of multistorey building is d = 25 m
Initial velocity of ball thrown upwards from point A,
` v _(0) = 20 ms ^(-1)`
Acceleration `a = g = - 10 ms ^(-2)`
Final velocity at point B, `v =0 ms ^(-1)`
Suppose, ball covers height h from A to B,
From third equation of motion,
`v ^(2) - v _(0) ^(2) =2ah `
`therefore 0- (20)^(2) = 2 (-10) xx h`
`therefore -400 =- 20 h`
`therefore h = 20 m`
Bal will rise upto 20 m from top building
(b) Total time of flight of ball can be find out in two parts,
(i)`t_(1)` is time taken to rise from A to B and
(ii)` t_(2)` is time taken to fall from B to C.
`therefore (i)` From first equation of motion.
`AB = 20 cm`
In `v = v _(0) + at, v = 0 and v _(0) = 20 m//s,`
`a =- 10 m//s^(2), 0 =20 -10 t _(1), t =t_(1)`
`therefore t _(1) = 2s" "...(1)`
(ii) Now `BC=20 + 25 = 45 m,` time taken to cover it is `t _(2),`
`S = v _(0) t 1/2 at ^(2)`
`S = BC = 45 m, a =- 10 ms ^(-2)`
` therefore - 45 = 0 xx t _(2) + (1)/(2) xx (-10) t _(2) ^(2)`
`therefore -45 =- 5t _(2) ^(2)`
`therefore t _(2) ^(.2) = 9`
`therefore t _(2) = 3s " "...(2)`
`therefore ` Total time of flight,
`t =t _(1) + t _(2)`
`t = 2 + 3`
[From equation (1) and (2)]
`t = 5s`
(b) Second method: Total time can be determined from initial final positions of ball with respect to origin.
From equation of motion,
`y =y _(0) + v _(0) t + 1/2 at ^(2)`
`y _(0) = 25 m, v _(0) = 20 ms ^(-1), y= 0 m `
`a =- 10 ms ^(-2)`
`therefore 0=25 + 20 t + 1/2 (-10 ) t ^(2)`
`therefore 5 t ^(2) = 25 + 20 t`
`therefore 5t^(2) - 20 t -25=0`
`therefore t ^(2) -4 t -5 =0`
`therefore (t -5) (t +1) =0`
`therefore t -5 =0 or t + 1 =0`
`therefore t =5 s or t =-1 s ` which is impossible.
`therefore t = 5 s`