A projectile is thrown with a velocity of 20 `m//s,` at an angle of `60^(@)` with the horizontal. After how much time the velocity vector will make an angle of `45^(@)` with the horizontal (in upward direction) is (take g=`10m//s^(2))-`

Options

A. `sqrt(3) sec`
B. `1//sqrt(3) sec`
C. `(sqrt(3-1) sec`
D. None of these

Answer

C

Explanation

As horizontal component of velocity remains unchanged

`"tan" 0=(V_(y))/(V_(x)) =(u "sin"60^(@) -"gt")/(0"cos" 60^(@) )=(20xxsqrt(3)/(2)-10xx1)/(20-1/2)`
`"tan" 45^(@) = (sqrt(3)/(2) -t/2)/((1)/(2))`
`1=(sqrt(3)-t)/(1)`
`t= (sqrt(3)-1)`