Physics - Kinematics Question with Solution | TestHub

The initial horizontal velocity is $u_x=u\cos {\theta }_0=20\cos 60^{\circ }=10m/s$.

The initial vertical velocity is $u_y=u\sin {\theta }_0=20\sin 60^{\circ }=10\sqrt{3}m/s$.

At time $t$, the horizontal velocity remains $v_x=u_x=10m/s$.

The vertical velocity is $v_y=u_y-gt=10\sqrt{3}-10t$.

When the velocity vector makes an angle $\theta$ with the horizontal, $\tan \theta =\frac{v_y}{v_x}$.

Given $\theta =45^{\circ }$, so $\tan 45^{\circ }=1$.

$1=\frac{10\sqrt{3}-10t}{10}$

$10=10\sqrt{3}-10t$

$10t=10\sqrt{3}-10$

$t=\sqrt{3}-1s$.

The final answer is $\boxed{(\sqrt{3}-1)\sec }$.