Question

A particle is projected with an initial velocity of 200 metres per second in a direction angle of 30° with the vertical the horizontal distance covered by the particle in 3 seconds is​

Answer 1

Answer:

300 m

Explanation:

Given :

Initial velocity ( u ) = 200 m / sec

Angle = 30

Time ( t ) = 3 sec.

We have to find Horizontal distance ( x ).

We know formula for x component. i.e.

x = u sin theta × t

We know sin 30 = 1 / 2

Now put the value in formula we get

x = 200 × 1 / 2 × 3 m

x = 100 × 3 m

x = 300 m .

Thus we get Horizontal distance 300 m.

Answer 2

Initial velocity of the projectile along horizontal direction, ux = u cos60 = 100 m/s

Initial velocity of the projectile along vertical direction, uy = u sin60 = 100

At the highest point, Velocity of the projectile along horizontal direction vx = 100m/s

Velocity of the projectile along vertical direction, vy = 0m/s

Momentum of the projectile along vertical direction, py = 0 kg m/s.  

Momentum of  particle A =  100m kg m/s (vertical upward direction)

Momentum of particle B = -100m kg m/s (vertical downward direction)

Momentum of particle C = mv'' kg m/s.  

Conservation of momentum in y direction

100m + (-100m) +mvy''= 0

vy'' = 0

Conservation of momentum in x direction

M100 = mvx''

vx'' = 100M/m

where m = M/3

vx'' = 300 m/s

The velocity of the particle is 300 m/s along the horizontal direction.  

NOTE: The velocity of the second particle is not mentioned in your question. In the answer it is assumed as the same velocity.