A student is able to throw a ball vertically to maximum height of 40m. The maximum distance to which the student can throw the ball in the horizonal direction:-
A. `40 (2)^(1//2)m`
B. `20(2)^(1//2)m`
C. 20m
D. 80m
Correct Answer - D
In verticle upwards. PE=mgh
`=mg((V_(2))/(2g))=(mv^(2))/(2)`...(i)
at an angle of `60^(@)` with vertical PE=mgh
`=mg((V^(2)sin^(2)30^(@))/(2g))=(mv^(2))/(8)`....(ii)
From (i)/(ii). `((PE)_(up))/((PE)_("at" theta=60^(@)"fromvt.))=((mv^(2))/(2))/((mv^(2))/(8))=(4)/(1)`
Correct Answer - D
Block B will come to rest when velocity of block A = velocity of block C.
`implies underset(0)overset(t)int12tdt=underset(0)overset(t)int3dt`
`6t^(2)=3t`
`t = 0, 0.5`
`therefore " "t =...
Correct Answer - B
`(b)` For the first place `5` players (excluding the oldest) and for the remaining place `5` (including the oldest)players are available.
`:.` No of ways `=5xx5xx4xx3xx2xx1=600`