A particle is projected from a point O with a velocity u at an angle `alpha` (upwards) to the horizontal. At a certain point P it moves at right angles to its initial direction. It follows that:

Options

A. OP makes an angle `tan^(-1) (u//2g)` to the horizontal
B. the distance of P from O is `u//(2g sin alpha)`.
C. the time of flight from O to P is `u//(g sin alpha)`
D. the velocity of the particle at P is u cot `alpha`

Answer

C, D

Explanation

Taking initial and final velocity on x and y axis.
0 = u - g sin `alphat`
`t =(u)/(g sin ^(2))`
`x = u [(u)/(g sin^(2))]-(1)/(2)(g sin alpha)[(u)/((g sin^(2)))]^(2)=`
`(u^(2))/(2gsinalpha)`
`y = (1)/(2)g cos a [(u)/(g sinalpha)]^(2)=(u^(2))/(2g)((cosalpha)/(sin^(2)alpha))`

OP = `sqrt(x^(2)+y^(2))=(u^(2))/(2gsinalpha)`
`tanphi=(y)/(x)=(cos alpha x sinalpha)/(sin^(2)alpha)= cot alpha`
`v = 0+g cos a [(u)/(g sin alpha)]= u cot alpha`