Question

6.
Equation of motion of projectile in x-y plane is
y = ax - bx?, then the angle of projection with
x-axis is
[NCERT pg. 78]
(1) tan (5)
(2) tan-'(a)
(3) cos-(b)
(4) sin-'(a)​

Answer 1

Equation of motion of projectile in x - y plane is y = ax - bx² ,

To find : The angle of projection with x - axis.

solution : method 1 : equation of motion of projectile is given as y = ax - bx²

on comparing with equation of projectile motion, y = x tanθ - gx²/2u²cos²θ

a = tanθ , b = g/2u²cos²θ

here tanθ = a ⇒θ = tan¯¹(a)

Therefore the angle of projection is tan¯¹(a).

method 2 : differentiating equation with respect to time.

dy/dt = a dx/dt - 2bx dx/dt

$v_y=a v_x-2bxv_x$

at initial time, x = 0, $v_x=u_x$ and $v_y=u_y$

so, $u_y=au_x-0$

⇒$\frac{u_y}{u_x}=a$

as $u_x=ucos\theta$, $u_y=usin\theta$

so, $tan\theta=a$ ⇒θ = tan¯¹(a)

Answer 2

Answer:

it is (2) tan-'(a)