Question

A particle is projected at an angle 30 with the horizontal with a velocity 10m/s then after 2s the velocity of particle

Answer 1

We can divide it into two components ucos theta and usin theta since the particle is in horizontal we will only use ucos theta which is equal 10 * 1/2 = 5.
We have to find v where we have been given u,t and a. v= u+at
v= 5 + 10(2)
v= 25
Hope it helps and if possible then please mark it as the brainliest answer.

Answer 2

Answer:after 1 sec the velocity of particle makes an angle of 60° with initial velocity vector

The magnitude of velocity of particle after 1sec is 10m/s

Explanation:

A PARTICLES IS PROJECTED AT AN ANGLE OF 30 WITH HORIZONTAL WITH A VELOCITY 10 m/s

INITIAL CONDITION

ux​ = 10cos 30° ay = -g

uy​ = 10sin 30° ax = ​0

after 2 seconds it would have already hit the ground so this case is not possible

the time of flight of the projectile is T = 2usin 30°/ g

T = [2 x 10 x (1/2)]/ 10

T = 1 sec.

vx = ux​ ​

and vy2 = uy2 + 2gy

as y = 0 after 1 sec

so vy =uy

hence it strikes the ground with the same velocity with which it was projected after 1 seconds and that is 10m/s

so the correct answer is c

and as it strikes in an identical manner as it was projected so it makes an angle of 30° with the horizontal and so the net angle which it makes with the initial vector is 30°+30° = 60° .

Hope it helps pls vote