IIT JEE , AIEEE Forum - Physics , Mechanics - PROJECTILE QUESTION:-

anshul_2009

A stone is thrown with velocity v at an angle B with horizontal. Find its speed when it makes an angle A with horizontal?

Plz . give the steps also .

Answer is : ( v cos B / cos A)

pottermania1990

since the horiz. vel remains a constant

let the vel at angle B be v'

vcosA = v'cosB

v' = vcosA/cosB

paddy.dude

yup potter is rght the horizontal component of velocity always remains constant in projectile motion

Prajju

Its simple.....
at any pt of the motion of the projectile direction = tan@ = Vy / Vx
now horizontal vel remains constant throughout the motion therefore Vx = VcosB
now tanA = Vy / VcosB
==> Vy = tanA.VcosB
now resultant vel. at any pt = sqrt ( Vy^2 + Vx^2 )
==> V' = sqrt ( V^2cos^2Btan^2A + V^2 cos^2B )
==> V' =VcosB sqrt ( 1 + tan^2A)
==> V' = VcosB secA
==> V' = VcosB / cosA

Prajju

Hope you are able to understand........and rate makes a lot of diff.

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