A ball is thrown vertically upwards from the top of a tower of height 60 m with a speed of 20 m//s.
(i) The ball strikes the ground after 6 s
(ii) The ball strikes the ground with speed `40 m//s`
(iii) The distance of ball above the ground after `5 s` is `35 m`
The maximum height attained by ball above the ground is 80 m

Options

A. (i), (ii)
B. (i),(ii),(iii)
C. (ii),(iv)
D. All option are correct

Answer

D

Explanation

`O`: origin
`O` to `A`: displacement`=-60 m`
`h=ut-1/2g t^(2)`
`-60=20t-5t^(2)`
`t^(2)-4t-12=0`
`(t-6)(t+2)=0`
`t=6 s`, (i) is correct
`v=u-g t`
`=20-10xx6=-40 m//s`, (ii) is correct
After `5 s`, displacement,
`h^(')=ut-1/2g t^(2)=20xx5-1/2xx10xx5^(2)`
`=100-125=-25 m`
-ve sign, i.e., ball is below `O` by `25 m`, hence distance of ball above the ground is
`60-25=35 m`, (iii) is correct
Maximum height of ball above `O`
`0=(20)^(2)-2xx10xxh_(0)impliesh_(0)=20 m`
Maximum height above ground `=60+20=80 m`, (iv) is correct.