A projectile is thrown in the upward direction making an angle of $${60^ \circ }$$ with the horizontal direction with a velocity of $$147\,m{s^{ - 1}}.$$  Then the time after which its inclination with the horizontal is $${45^ \circ },$$  is

Solution :
At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then $$u\cos {60^ \circ } = v\cos {45^ \circ }.$$
$$\eqalign{ & \Rightarrow 150 \times \frac{1}{2} = v \times \frac{1}{{\sqrt 2 }}\,\,{\text{or}}\,\,\frac{{150}}{{\sqrt 2 }}m/s \cr & {\text{Initially}}:\,{u_y} = u\sin {60^ \circ } = \frac{{150\sqrt 3 }}{{\sqrt 2 }}m/s \cr & {\text{Finally}}:{v_y} = v\sin {45^ \circ } = \frac{{150}}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = \frac{{150}}{2}m/s \cr & {\text{But}}\,\,{v_y} = {u_y} + {a_y}t\,\,{\text{or}}\,\,\frac{{150}}{2} = \frac{{150\sqrt 3 }}{2} - 10t \cr & 10t = \frac{{150}}{2}\left( {\sqrt 3 - 1} \right)\,{\text{or}}\,t = 7.5\left( {\sqrt 3 - 1} \right)s \cr} $$