Question

At time $t=0$, a small ball is projected from point $A$ with a velocity of $60m/s$ at ${60}^o$ angle with horizontal. Neglect atmospheric resistance and determine the two times $t_1$ and $t_2$ where the velocity of the ball makes an angle of ${45}^o$ with the horizontal $x-$ axis.

Answer 1

at both point , $v_x=u\cos {60}^0=30m/s$

$v_y=u\sin {60}^0-gt=30\sqrt{3}-10t$

when , velocity makes ${45}^0$ then $\tan {45}^0=+1or-1=v_y/v_x$

$30\sqrt{3}-10t=30\Rightarrow t_1=3(\sqrt{3}-1)s$

Or, $30\sqrt{3}-10t=-30\Rightarrow t_2=3(\sqrt{3}+1)s$