Question

A projectile is projected with initial velocity of 30 m/s 3 at an angle 8 = tan -1 After 1 second, direction of motion of the particle makes an angle a with horizontal, then angle a is given as (g = 10 m s-2)​

Answer 1

Answer = 1/3

Answer 1/ 3

Answer 1/3

Answer 2

Given,

u = 30 m/s

angle, tan = 1/8

To find,

At t = 1 sec find measure of angle a

Solution,

tanx = 1/8

sinx = 1/√65

cosx = 8/√65

Along direction of motion after t = 1 sec

v = u + at

vsina = usinx + (-g)t

vsina = 30 × 1/√65 - 10

vsina = 30/√65 - 10

vsina = (30 - 10√65)/√65                                                       (i)

Horizontal component will remain constant throughout the journey

vcosa = ucosx

vcosa = 30 × 8/√65                                                                (ii)

Dividing equation i and ii

$\frac{vsina}{vcosa} = \frac{(30 - 10\sqrt{65})}{\sqrt{65}}$ × $\frac{\sqrt{65} }{240}$

sina/cosa = (30 - 10√65) / 240

tana = (30 - 10√65) / 240

a = tan⁻¹ (30 - 10√65) / 240

Angle a is given as tan⁻¹ (30 - 10√65) / 240.