IIT JEE , AIEEE Forum - Physics , Mechanics - what is magnitude of initial velocity & angle of projection

rakesh61

After one second the velocity of projectile makes an angle of 45⁰ with horinzotal. After another one second it is travelling horizontally. The magnitude of its intial velocity and anglen of projection are (g=10m/ s^2)

ans 22.36 m/sec,tan^-1 (2)

please give detal answer

shubham_sachdeva

i think qn. is wrong ... angle should be 45 degrees..

also if u put the angle given in ans. then u will not get u right...

nishantsingh89

@shubham

the answer given is correct

let M_X,M_Y,BE velocity after 1 second of projection(when angle is 45 degree)
at that instant tan45 = 1
thus M_{X =}MY

the velocity along y axis is zero 2 seconds after projection

let V_ybe velocity along y axis at time of projection
at t=2 seconds velocity along y axis =0
0= V_y- 10*2
V_y=20 m/s
after 1 second of projection velocity along y axis will be

M_y = 20 - 10*1
M_{y = 10m/s
Also along x axis motion is unacclerated so velocity along x axis is always constant
Mx = 10 m/s = Vx (Vx initial projection velocity along x axis

velocity of projection V= sqrt( Vy^2 + Vx^2)
= sqrt(500)
= 22.36 m/s
angle of projection
tan = Vy/Vx = 2}

= tan^-1(2)

do rate me ..............if helpful :)

shubham_sachdeva

@nishant...

let ur answer be correct....

after 2 seconds velocity has only x direction....

hence it is T/2 = 2 sec..

T = 2usin theta/g

T/2 = u sin theta/g

tan theta = 2

so sin theta = 2/rt.3

so putting values given in answer we didn't get lhs = rhs..

anandghegde

download the image for clarity

nishantsingh89

@shubham

tan theta= 2

sin theta=2/rt.3

don't u think]
sin theta = 2/rt5???????????

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