A particle is projected from the ground with velocity u making an angle thetawith the horizontal. At half of its maximum heights,

A particle is projected from the ground with velocity u making an angl

1.	A particle is projected from the ground with velocity u making an angle thetawith the horizontal. At half of its maximum heights,
Answer» its HORIZONTAL velocity is `U cos THETA ` its vertical velocity is `(u SIN theta)/(SQRT2)` its velocity is `u((1+ cos^2 theta)/(2))^(1/2)` all the above are true Answer :D