Correct Answer - B::C
`u_x = 20 km//h = 20 xx 5/18 = 5.6 m//s`
`u_y = 12 km//h`
`= 12 xx 5/18 = 3.3 m//s`
Using `S = ut + 1/2 at^2` in vertical direction, we have
`-50 = (3.3) t + 1/2 (-10) t^2`
Solving this equaiton we get,
`t = 3.55 s`
At the time of striking with ground,
`v_x = u_x = 5.6 m//s`
`v_y = u_y + a_yt`
`= (3.3) + (-10) (3.55)`
`=32.2 m//s`
`:. Speed = (sqrt(32.2)^2 + (5.6)^2)`
` ~~ 32.7 m//s`.