A ball is thrown with a velocity u making an angle 'θ' with the horizontal. Its velocity vector is normal to initial velocity vector u after a time interval of
usinθg
ugcosθ
ugsinθ
4cosθg
A
4cosθg
B
ugcosθ
C
usinθg
D
ugsinθ
Open in App
Solution
Verified by Toppr
Initial velocity, u u=ucosθ^i+usinθ^j ux= constant
ux(t)=ucosθ uy(t)=usinθ−gt So, u(t)=ucosθ^i+(usinθ−gt)^j
Now, for u(t) to be perpendicular to u, dot product should be 0. →u(t)⋅→u=0
(ucosθ^i+(usinθ−gt)^j)⋅(ucosθ^i+usinθ^j)=0
⇒u2cos2θ+u2sin2θ−usinθgt=0
t=ugsinθ
Was this answer helpful?
17
Similar Questions
Q1
A ball is thrown with a velocity u making an angle 'θ' with the horizontal. Its velocity vector is normal to initial velocity vector u after a time interval of
View Solution
Q2
A particle is projected from the ground at an angle of θ with the horizontal with an initial speed of u. Time after which velocity vector of the projectile is perpendicular to the initial velocity is
View Solution
Q3
A ball is projected with velocity u making an angle θ with the horizontal. The magnitude of velocity at the instant when it is perpendicular to its initial velocity of projection