A projectile moves from the ground such that its horizontal displacement is `x=Kt` and vertical displacement is `y=Kt(1-alphat)`, where K and `alpha` are constants and t is time. Find out total time of flight (T) and maximum height attained `(Y_"max")`

Options

A. `T = alpha, Y_("max") = K/(2alpha)`
B. `T =1/alpha, Y_("max") =(2K)/alpha`
C. `T =1/alpha, Y_("max") = K/(6alpha)`
D. `T = 1/alpha, Y_("max") =K/(4alpha)`

Answer

D

Explanation

During time of flight vertical displacement becomes zero
i.e. y =0
`KT(1-alpha T) = 0 rArr T = 1/alpha`
Time taken by particle to attain max. height
`t = T/2 = 1/(2alpha)`
`y_("max") = K.1/(2alpha) (1- alpha.1/(2alpha)) rArr y_("max") = K/(4alpha)`