Two particles are simultaneously thrown at an elevation 45°, towards each other from points A and B , roofs of two high buildings. Their velocities of projection are 14 m/s and 2 m/s respectively. Hor

Question

Two particles are simultaneously thrown at an elevation 45°, towards each other from points A and B , roofs of two high buildings. Their velocities of projection are 14 m/s and 2 m/s respectively. Horizontal and vertical separation between points A and B is 22 m and 9 m respectively. Calculate the minimum separation between the particle in the process of their motion.

TestHub · Accepted Answer

Both particles have the same acceleration $g$ , so their relative motion is uniform. With $A$ at the origin and $B$ at $(22, 9)$ :
$v_{1} = (72, 72)$
and
$v_{2} = (- 2, 2)$ ,
so
$u = v_{2} - v_{1} = (- 82, - 62)$ ,
magnitude
$u = 102$ .
Minimum separation = (cross term)/(magnitude of $u$ ), with cross term
$22 \times (- 62) - 9 \times (- 82) = - 602$ :
$\frac{60 2}{10 2} = 6$
m. Matches the printed answer. ( $B$ is the higher roof; with $B$ lower the result would not match.)

Physics - Kinematics Question with Solution | TestHub

Doubts & Discussion