Question

Two particles are projected horizontally from a tower with velocity 3m/s and 4m/s respectively but in opposite direction. After what time their velocities will make an angle 90 deg. With each other?
t = 1 s
t = root 0.12 s
t = root 2 s
t = 2 s

Answer 1

Trick here is to write the velocities in component form.
Let $(3,v_1)$ be the velocity of first particle.
Let $(-4,v_2)$ be the velocity of second particle.

Since the vertical component of initial velocity is $0$ for both the particles, we have $v_1=v_2=gt=10t$

We want to know the time at which the velocities are perpendicular, so simply set the dot product of velocities equal to $0$  and solve $t$ : 

$(3,10t)\cdot~(-4,10t)=0\implies -12+100t^2=0\implies~t=\sqrt{0.12}s$

Answer 2

Answer:T= √0.12s

Explanation:

Let ( 3, v1) be first velocity

Let (-4,v2)be second velocity

So,

V1 = v2= get = 10t

(3,10t) * (-4,10t) =0

-12 + 100t^2 = 0

t = √ 0.12s