Question

A projectile is thrown at an angle of 60 with the horizontal. After how much time will its inclination with the horizontal be 45 ? ( Given V= 147m/s) .

Answer 1

use formula

tanB=(UsinA-gt)/UcosA
where B is final inclination of velocity with horizontal , A is initial inclination of velocity with horizontal and U is velocity with projectile projected . and t is time .
now
t=(UsinA-UcosA.tanB)/g
=147 (sin60-cos60.tan45)/g
=147 (root3-1)/2g
=147 x (0.732)/20
=5.38 sec