A particle is projected at an angle `60^@` with horizontal with a speed `v = 20 m//s`. Taking `g = 10m//s^2`. Find the time after which the speed of the particle remains half of its initial speed.

Answer

C

Explanation

`v_x = u_x = 20 cos 60^@ = 10 m//s`
Given, `v=u/2`
`:. 4 v^2 = u^2`
or `4(v_(x)^2 + v_(y)^2) = u^2`
`:. 4[(10)^2 + v_(y)^2] = (20)^2`
or `v_y = 0`
Hence, it is the highest point
`:. t = T/2 = (usintheta)/g`
`=((20)sin60^@)/10`
`=(sqrt3) s`.