Question

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

Answer 1

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

Answer 2

`v_0=sqrt(v_x^2+(v_y+g t)^2`
`Deltay=v_yt+1/2g t^2`
where `(v_x,v_y)=(12,-4)m/s;, t=2s`

Explanation:

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.