Question

the direction of a motion of a projectile at a certain instant is inclined at an angle a to the horizontal. After t second inclined at an angle b. find the horizontal component of velocity of projection in term of g,t,a and b (a and b are positive angle in anticlockwise direction​

Answer 1

Let a projectile is projected with speed u at an angle ∅ with horizontal.

so, horizontal component of initial velocity, ux = ucos∅.......(1)

at certain instant t' , projectile is inclined at an angle a to the horizontal.

so, vertical component of projectile at that instant, vsina = usin∅ - gt' .....(2)

and horizontal component of projectile at that instant , vcosa = ucos∅ ....(3)

then from equations (2) and (3),

usina.cos∅/cosa = usin∅ - gt' .....(4)

similarly,

now after time t, inclined at angle b.

so, vertical component of projectile, v'sinb = usin∅ - g(t' + t)

and horizontal component of projectile, v'cosb = ucos∅

then, usinb.cos∅/cosb = usin∅ - gt' - gt.....(5)

from equations (4) and (5),

ucos∅/cosb = ucos∅/cosa - gt

or, gt = ucos∅ [ sina/cosa - sinb/cosb ]

= ucos∅ [ tana - tanb ]

so, ucos∅ = gt/[tana - tanb]

hence, horizontal component of projectile is gt/[tana - tanb]