Question

2.
A particle is projected at an angle of elevation
sin-(4/5)
and it's range on the horizontal plane is
3 km. The Initial velocity of projection is
(1) 160 m/s
(2) 175 m/s
(3) 180 m/s
(4) 230 m/s​

Answer 1

Answer:

Option (2) 175 m/s

Explanation:

The angle of projection of the particle = $\sin^{-1}(4/5)$

If the angle of projection is θ

Then

$\theta=\sin^{-1}(4/5)$

or $\sin\theta= 4/5$

Therefore, $\cos\theta =\sqrt{1-\sin^2\theta}= \sqrt{1-(4/5)^2}=\sqrt{9/25} =3/5$

The range R of the projectile = 3 km = 3000 m

using the range formula

$\boldsymbol{R=\frac{u^2\sin2\theta}{g}}$, where u is the initial velocity

we get

$R=\frac{u^2\times 2\sin\theta.\cos\theta}{g}$    (∵ sin2θ = 2sinθcosθ)

or, $u^2=\frac{R\times g}{2\sin\theta\cos\theta}$

$\implies u^2=\frac{3000\times 9.8}{2\times\frac{4}{5}\times\frac{3}{5}}$

$\implies u^2=\frac{300\times 98\times 25}{2\times 4\times 3}}$

$\implies u^2=25\times 49\times 25$

$\implies u=5\times7\times 5=175$

Therefore the initial velocity of the projection is 175 m/s