A long lorry is moving with constant velocity along a horizontal road. Another identical lorry is moving ahead of it with the same velocity. A small object is projected from a point with position vector $\vec r_1$ on the first lorry at an angle of $45^\circ$ above the horizontal. What should be the horizontal-plane component of the object's velocity relative to the first lorry so that the object reaches a point with position vector $\vec r_2$ on the second lorry? Neglect air resistance.
Solution
Since both lorries have the same constant velocity, the second lorry is at rest relative to the first. Thus, in the first-lorry frame, the projectile has to cover the horizontal displacement
$$\vec d=\vec r_2 - \vec r_1$$
Its horizontal range is
$$R=|\vec r_2 - \vec r_1|$$
For a projectile launched at $45^\circ$,
$$R=\frac {u^2sin90^\circ}{g}=\frac {u^2}{g}$$
$$u=\sqrt {g|\vec r_2 - \vec r_1|}$$
The horizontal-plane velocity vector is therefore in the direction from $\vec r_1$ to $\vec r_2$, i.e.
$${{\vec u}_H} = u\cos {45^\circ }\frac{{{{\vec r}_2} - {{\vec r}_1}}}{{\left| {{{\vec r}_2} - {{\vec r}_1}} \right|}}$$
$${{\vec u}_H} = \sqrt {\frac{{g\left| {{{\vec r}_2} - {{\vec r}_1}} \right|}}{2}} \frac{{{{\vec r}_2} - {{\vec r}_1}}}{{\left| {{{\vec r}_2} - {{\vec r}_1}} \right|}}$$
or, equivalently,
$${{\vec u}_H} = \sqrt {\frac{g}{{2\left| {{{\vec r}_2} - {{\vec r}_1}} \right|}}} ({{\vec r}_2} - {{\vec r}_1})$$
Note: Since, the lorries are identical, the two points are at the same vertical level.