Question

Two particles are projected horizontally and simultaneously from top of a tower in mutually perpendicular planes with same speed 30m/s. After how much time their velocity vectors will be at angle 60° from each other?

Answer 1

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Answer 2

After 4 seconds the velocity vector of the particle will be at 60°.

Given:

Speed  of particles = 30m/s.

Angle the particles make = 60°

To Find:

The time at which they will make 60° angle.

Formula Used:

m =  $\frac{u sin60 - gt}{v cos60}$

Explanation:

Let us write their velocities  in vector form for a particle projection with same velocity u at θ

The velocity is given as

The slope will be

m =  $\frac{u sin60 - gt}{v cos60}$                                            (a)

For velocities becoming parallel

Substituting the values in equation (a) we get

$\frac{u sin60 - gt}{v cos60}$

(u(√3/2) - gt))/ (1/2)

((u/2) - gt)/(√3/2) = (u(√3/2) - gt))/ (1/2)

(u/2)- gt = (3u/2) - √3gt

(√3 + 1)gt = (3u/2) - (u/2)

So now

t = 30/((√3 - 1) × 10)

 = 3(√3 + 1) / (3 - 1)

t = 3( 1.7 + 1)/(√3)² - (1)²

 = $\frac{3( 2 - 7)}{2}$

 = $\frac{8.1}{2}$

 = 4 seconds

Time = 4 seconds.