Question

A projectile is thrown in the upward direction making an angle of `60^@` with the horizontal direction with a velocity of `150 ms^-1`. Then the time after which its inclination with the horizontal is `45^@` is

Answer 1

Here, ` theta= 60^@, u =150 ms^(-1)` . Let after time (t) its velocity become (v) making an angle ` 45^@` with the horizontal. Since the horizontal component velocity of projectile remains sunchanged during angular projection, henc
` u cos 60^@ = v cos 45^2` or ` v= ( u cos 60^@0/ ( cos 45^@)`
Taking vertcal upward motin of projectile from tiem (t), we have, ` u =u sin 60^@, a =- g, t=t`
and ` v=v sin 45^@`. ltbRgt As ` v=u + at so v sin 45^2 = u sin 60^@- gt`
or ` ( u cos 60^@)/( cos 45^@) xx sin 45^@ = u sin 60^@-gt` ltbRgt or ` t = u/g [ sin 60^@- cos 60^@] = ( 150)/(10)[(sqrt3)/2 - 1/2]` ltbRgt =` 5.49 s`.