Question

Two bodies were thrown simultaneously from the same point: one, straight up, and the other, at an angle of $\theta ={60}^{\circ }$ to the horizontal. The initial velocity of each body is equal to $v_0=25m/s$. Neglecting the air drag, the distance(in meters) between the bodies $t=1.70s$ later is (20+x). The value of x is :

Answer 1

The solution of this problem becomes simple in the frame attached with one of the bodies. Let the body thrown straight up be 1 and the other body be 2, then for the body 1 in the frame of 2 from the kinematic equation for constant acceleration:
$\overrightarrow{r_{12}}={\overrightarrow{r_0}}_{\left(12\right)}+{\overrightarrow{v_0}}_{\left(12\right)}t+\frac{1}{2}{\overrightarrow{w}}_{12}{t}^2$
So, $\overrightarrow{r_{12}}={\overrightarrow{v_0}}_{\left(12\right)}t$, (because ${\overrightarrow{w}}_{12}=0$ and ${\overrightarrow{r_0}}_{\left(12\right)}=0$)
or $|\overrightarrow{r_{12}}|=|{\overrightarrow{v_0}}_{\left(12\right)}|t$ (1)
But $|{\overrightarrow{v}}_{01}|=|{\overrightarrow{v}}_{02}|=v_0$
So, from properties of triangle
$v_{0\left(12\right)}=\sqrt{v_0^2+v_0^2-2v_0{v}_0\cos (\frac{\pi }{2}-{\theta }_0)}$
Hence, the sought distance $|\overrightarrow{r_{12}}|=v_0\sqrt{2(1-\sin \theta )}t=22m$