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A particle is projected from a point P with a velocity v at an angle θ with horizontal. At a certain point Q it moves at right angles to its initial direction. Then (a) velocity of particle at Q is vsintheta (b) velocity of particle at Q is Vtantheta (c) time taken from P and Q is v/g sin theta (d) time taken from P and Q is v/gsectheta
Asked by hy9022075 | 11 Jan, 2023, 05:06: PM
Figure shows the particle projected at P with velocity v at angle θ with horizontal.

When the particle reaches the point Q in its trajectory its direction of velocity is perpendicular to its intial velocity
as shown in figure.

Let u be the velocity of particle at Q . Its horizontal velocity component = u cosφ = u cos(90-θ) = u sinθ

if v is inital velocity at P , then horizontal component v cosθ remains constant throughout its trajectory.

Hence , we have ,  u sinθ = v cosθ

Hence velocity at Q is given as,

u = v ( cosθ / sinθ ) = v cotθ  ...................................(1)

At Q , vertical component of velocity ,  u sinφ = u sin(90-θ) = u cosθ = v cotθ cosθ

Time taken t to reach Q from P is determined from velocity at both position.

v cotθ cosθ = v sinθ - (g t )

from above expression we get

t = ( v / g ) [ sinθ - cotθ cosθ ]

Answered by Thiyagarajan K | 12 Jan, 2023, 12:02: AM

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