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

2 Answers
Feb 9, 2018

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

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

So,suppose,after 2s2s of its journey, it had a vertical velocity of vv so we can write, v= u sin theta -g *2v=usinθg2 (using, v=u-g tv=ugt,here, initial velocity upwards is u sin thetausinθ)

So,in vector form we can write,velocity after 2s2s as,

(u cos theta)i + (u sin theta -2g)j(ucosθ)i+(usinθ2g)j

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

u cos theta =12ucosθ=12 and u sin theta -2g=-4usinθ2g=4

Solving the above two equations we get, u=20 m/su=20ms and theta = cos ^-1(3/5)=53θ=cos1(35)=53 degrees

Feb 9, 2018

v_0=sqrt(v_x^2+(v_y+g t)^2v0=v2x+(vy+gt)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.