1

Question

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

Open in App

Solution

The correct options are

**C** velocity of particle at Q is vcotθ

**D** time of flight from P to Q is vg cosec θ

Let net velocity at Q is vq, it makes an (θ−90) with horizontal as it is vertical to vq and vp makes θ with horizontal .

Now as there is no acceleration along x and air resistance is ignored , horizontal component of velocity remains unchanged .

i.e vpcosθ = vqcos(θ−90)

vq = vpcotθ = vcotθ ..... (i)

initial vertical velocity +vsinθ

Final vertical component of velocity

= vqsin(θ−90) =−vcotθ×cosθ = −v(cosθ)2/sinθ

Using laws of motion :

−v(cosθ)2sinθ=vsinθ−gt

t=vcosecθ/g .... (ii) (Time of flight from P to Q )

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program