Question

A particle is projected with velocity u at angle theta with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

Answer 1

time t = u/gsinθ when velocity is perpendicular to its initial velocity.

it is given that a particle is projected with velocity u at an angle θ with the horizontal.

we have to find the time when velocity vector is perpendicular to its initial velocity vector.

initial velocity vector, $\vec u=ucos\theta\hat i+usin\theta\hat j$

after time t,

velocity vector will be, $\vec v=ucos\theta\hat i+(usin\theta-gt)\hat j$

now $\vec v$ and $\vec u$ are perpendicular to each other.

so, $\vec v.\vec u=0$

⇒$(ucos\theta\hat i+usin\theta\hat j).\{ucos\theta\hat i+(usin\theta-gt)\hat j\}=0$

⇒u²cos²θ + u²sin²θ - ugsinθ t = 0

⇒u²(sin²θ + cos²θ) - ugsinθt = 0

⇒u² = ugsinθt

⇒t = u/gsinθ

Answer 2

Answer:time t = u/gsinθ when velocity is perpendicular to its initial velocity.

it is given that a particle is projected with velocity u at an angle θ with the horizontal.

we have to find the time when velocity vector is perpendicular to its initial velocity vector.

initial velocity vector,

after time t,

velocity vector will be,

now and are perpendicular to each other.

so,

⇒

⇒u²cos²θ + u²sin²θ - ugsinθ t = 0

⇒u²(sin²θ + cos²θ) - ugsinθt = 0

⇒u² = ugsinθt

⇒t = u/gsinθ