After 2 seconds of flight,the velocity of a projectile is (12i-4j)m/s,Find its speed and elevation at moment of projection?

When a projectile is projected with an initial speed of #u# at an angle of #theta# w.r.t to horizontal,its horizontal component of velocity i.e #u cos theta# remains constant through out.

But, the vertical component of velocity i.e #u sin theta# decreases to zero,while going up,as gravitational force opposes that motion.

So,suppose,after #2s# of its journey, it had a vertical velocity of #v# so we can write, #v= u sin theta -g *2# (using, #v=u-g t#,here, initial velocity upwards is #u sin theta#)

So,in vector form we can write,velocity after #2s# as,

#(u cos theta)i + (u sin theta -2g)j#

So,comparing with the given equation,we can say,

# u cos theta =12# and #u sin theta -2g=-4#

Solving the above two equations we get, #u=20 m/s# and #theta = cos ^-1(3/5)=53# degrees

Let's set t=0 at the moment of projection, and denote
#v_0, v_(0x), v_(0y)# as initial velocity and initial velocities in x- and y- direction.

#(v_x, v_y) = (12, 4)m/s#

#because v_y=v_(0y)- g t#
#rArr v_(0y)= v_y+g t#

Substitute the above into the y(t) equations

#y=y_0+v_(0y)t-½g t^2#
#Deltay=(v_y+g t)t-1/2g t^2=v_yt+1/2g t^2#

Because,
#v_(0x)=v_0costheta=v_x#
#v_(0y)=v_0sintheta=v_y+g t#

#rArr v_0= sqrt(v_(0x)^2+v_(0y)^2)=sqrt(v_x^2+(v_y+g t)^2#

Substitute #v_x = 12 m/s, v_y= -4m/s, and t=2s#, you should be able to calculate the elevation, and the initial velocity.