Question

1. Two particles are projected from a tower horizontally in opposite directions with velocities 10 m/s
and 20 m/s. Find the time when their velocity vectors are mutually perpendicular. Take
g=10m/s2​

Answer 1

Answer:

The time is 18 m

Explanation:

According to the problem two particles are projected  horizontally

let the velocity of the  first particle x direction is v1,

Therefore, v1=10 i m/s

and  the velocity of the other particle  is v2

Therefore, v2= -20 i m/s

Let the time when their  velocity vectors are mutually perpendicular is t,

Therefore,

s=v+a*t.

as we know  as no force acts on horizontal,

Therefore the  vertical velocity of the first body is  v1=0, a=-10m/s, s1= -10t m/s

and v2 also equal to -10t m/s  

d1= 10i – 20t j,    d2= -20i -10t j;

after solving we get, t=0.6 sec,

for the first particle the time taken,

tx=10 x 0.6 = 6

ty=-20x a x t =-1.8 m

so S1=6i-1.8j

S2= 12i-1.8j

therefore the time between them 6+12=18 m

Answer 2

Answer:

Answer is √2 seconds

Explanation:

As particle is thrown horizontally Initial velocity will only be in X axis direction

also, both are thrown opposite, so one will be in +ve X axis nd 2nd will be in -ve X axis direction

Let both Vector velocity will be perpendicular after 't' seconds.

Final velocity of 1st particle = u+at == 10i+(-10j)t == 10i-10jt

[Taking acc. j as it is only effective in y axis direction or downwards direction] ---- {acc due to gravity}

Final velocity of 2nd particle = u+at == -20i+(-10j)t == 20i-10jt

[here I'm taking 'u' as -ve bcz both were thrown opposite along horizon]

Since, both vectors are Perpendicular, their dot product will be 0

So, V1 . V2 = 0

= (10i-10jt) . (-20i-10jt) = 0

= -200+100t^2 = 0

= t^2 = 200/100

= t = √2 sec.