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

let v velocity , when inclination goes to 45 with horizontal .
also velocity along x axis remains constant
so,
Ucos60=Vcos45
V=Ucos60/cos45=147cos60/cos45

hence Vy=Vsin45
=147cos60.tan45=147cos60

equation for y component
Vy=Uy+ayt
147cos60=147sin60-gt
t=(147sin60-147cos60)/g=147 (0.732)/20
=5.3802 sec

Answer 2

u=147m/s      initial angle = 60°
u_x = 147 cos 60° m/s = 73.5 m/s
For the inclination of the velocity to be 45° with the horizontal, the vertical velocity u_y should be same as the horizontal velocity u_x, ie., 73.5 m/s
Apply the equation of motion with g = 10 m/s²

73.5 = 147 * sin 60° - 10 * t
t = 73.5 (√3 -1)/10 = 5.38 sec