A particle is projected upwards with a velocity of 100m/s at an angle of 60o, with the vertical. Then time taken by the particle when it will move perpendicular to its initial direction-(A) 10 sec(B) 20/ root3 sec(C) 5 sec(D)10 root3 sec

Prithvi Raj
34 Points
9 years ago
I think options given are wrong,
Lets consider a normal projectile,
the initial velocity makes 60 degree so ve can make two components of it wrt to ground which are “u cosA” horizontally and “u sinA” vertically. now at some point of its trajectory the bodies new velocity makes angle 90 degrees to the intial velocity so at that point lets consider it makeing an angle B with ground (you can imagin that only after that projectile passess highest point it will be able to make 90 degree with intial velocity simply draw a figure) now lets make two components for our new velocity (v), which are “v cos B” horizondally and “v sinB” vertically. Now you have to think....  you know that vectors can be moved in a plane with out changing its angle and magnitude so drag our new velocity vector “v” and join its tail to intial velocity “u” now  “u” is making angle A=60 with ground and it also makes angle 90 with “v”,  from that figure “v” is making angle 30 degree with ground.
So now analyze 2D motion as two 1D motion along ground and perpendicular to ground.
Along ground,
v=u+at
v cosB=u cosA + 10 * t  since no acceleration along horizondal
from here u=100, A=60, B=30
so v = 100/root3
then vertical motion,
v sinB=u sinA – g*t
v=100/root3, B=30, u=100, A=60, g=10
so we get t=10/root3
Rishabh
28 Points
7 years ago
Initial velocity in vertical direction is 100*sin60°=50√3m/sInitial velocity in horizontal direction is100*cos60°=50m/sWhen particle is perpendicular to its initial direction then direction of its velocity will be -30° fro horizontaltan(-30°)=Vy/Vx=-1/√3Vy/50=-1/√3Vy=-50/√3Since v=u+at-50/√3=50√3-gt-50/√3=50√3-10tt=20/√3