Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and move with velocities v1 = 3.0 m/s and v2 = 4.0 m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Question

Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and move with velocities v1 = 3.0 m/s and v2 = 4.0 m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular. (A) 5 m (B) 7 √3 m (C) (7√3/5)m (D) (7/2)m

Tardigrade Admin · Accepted Answer

Let the particles move perpendicular to each time

t

.

Hence

(4 \hat{i} - g t \hat{j}) . (- 3 \hat{j} - g t \hat{j}) = 0

\Rightarrow - 12 + g^{2} t^{2} = 0

\Rightarrow t = (\frac{12}{1000}) \Rightarrow t = \frac{3}{5}

Hence distance,

d = (u t + 3 t^{2}) + ((h - \frac{1}{2} g t^{2}) - (h - \frac{1}{2} g t^{2}))^{2}

= 7 t = \frac{7 3}{5} m

$5 m$

$73 m$

$\frac{7 3}{5} m$

$\frac{7}{2} m$

5m

73​m

573​​m

27​m

Let the particles move perpendicular to each time t. Hence (4i^−gtj^​).(−3j^​−gtj^​)=0 ⇒−12+g2t2=0 ⇒t=(100012​)​⇒t=53​​ Hence distance, d=(ut+3t2)+((h−21​gt2)−(h−21​gt2))2​ =7t=573​​m

$5 m$

$73 m$

$\frac{7 3}{5} m$

$\frac{7}{2} m$