Question

a particle is projected at an angle of 30 to the horizontal and 2 sec later is moving in direction tan -1 (1/4) to the horizontal for the first time . find its initial speed​

Answer 1

Let u is the initial speed of the particle.

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

= $ucos30^{\circ}\hat{i}+usin30^{\circ}\hat{j}$

= (√3/2)u i + (1/2)u j

velocity of particle after t = 2s, $\vec{v}=ucos\theta\hat{i}+(usin\theta-gt)\hat{j}$

= $\vec{v}=ucos30^{\circ}\hat{i}+(usin30^{\circ}-2g)\hat{j}$

= (√3/2)u i + (1/2u - 2g)j .......(1),

a/c to question,

particle makes an angle tan-¹(1/4) with horizontal after first two 2sec.

so, v = vcos(tan-¹(1/4)) i + vsin(tan-¹(1/4)) j

= vcos(cos-¹(4/√17)) i + vsin(sin-¹(1/√17))

= (4/√17)vi + (1/√17)v j ........(2)

on comparing equations (1) and (2),

(√3/2)u = (4/√17)v

or, v = (√51/8)u .....(3)

(1/2u - 2g) = (1/√17)v

from equations (3),

(1/2u - 2g) = (1/√17) × (√51/8)u

or, 1/2u - 2g = (√3/8)u

or, (1/2 - √3/8)u = 2g

or, u = 16g/(4 - √3)