Question

Two particles thrown horizontally in opposite direction from the same point from height h simultaneously with velocity for metre per second and 3 metre per second the time taken when their velocity are perpendicular is approximately

Answer 1

Answer:

√3/5  =  0.35 sec

Explanation:

Two particles thrown horizontally in opposite direction from the same point from height h simultaneously with velocity 4 metre per second and 3 metre per second the time taken when their velocity are perpendicular is approximately

Let say time = t sec

a = g = 10 m/s²

using V =U + at

Horizontal Velocity = 4 m/s

Vertical Velocity at  t = 0 + 10t = 10t

Angle with Vertical  = Tanα  = 4/10t  = 2/5t

Horizontal Velocity = 3 m/s

Vertical Velocity at  t = 0 + 10t = 10t

Angle with Vertical  = Tanβ  = 3/10t

α + β = 90 => β = 90-α

=> Tan(90-α) = 3/10t

=> Cotα = 3/10t

Tanα * Cotα  = (2/5t) *(3/10t)

=> 1 = 3/25t²

=> t²  = 3/25

=> t = √3/5  =  0.35 sec