Two particles are simultaneously thrown at 45° elevation towards each other from the roofs of two tall buildings with speeds 7 2 m / s and 2 m / s . The horizontal and vertical separations between

Question

Two particles are simultaneously thrown at 45° elevation towards each other from the roofs of two tall buildings with speeds $72 m / s$ and $2 m / s$ . The horizontal and vertical separations between the roofs are 8 m and 11 m (the second roof higher). The minimum separation between the particles during the motion is:

Answer:

4

Solution:

$u = (- 1 - 7, 1 - 7) = (- 8, - 6)$ ,
magnitude
$u = 10$ ;
cross term
$8 \times (- 6) - 11 \times (- 8) = 40$ ,
so min separation =
$\frac{40}{10} = 4$
m.

TestHub · Accepted Answer

u = ( − 1 − 7 , 1 − 7 ) = ( − 8 , − 6 ) , magnitude u = 10 ; cross term 8 × ( − 6 ) − 11 × ( − 8 ) = 40 , so min separation = 10 40 ​ = 4 m.

Physics - Kinematics Question with Solution | TestHub

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