Question

A particle is projected at an angle of elevation
sin-1
and it's range on the horizontal plane is
3 km. The initial velocity of projection is.​

Answer 1

sorry for thatction is.

Answer 2

A particle is projected at an angle of elevation sin^ -1(4/5) and it's range on the horizontal plane is 3 km. The initial velocity of projection is

The Horizontal range is the Displacement of the projectile as it reaches the plane of projection.

It is given,

$\huge \: R = \frac {u ^{2} sin2 \theta }{g }$

Given,

Angle of projection θ = sin^-1( 4/5)

⇒sin θ = 4/5

⇒cosθ = 3/5

Range R = 3 * 10^3 m.

Now,

$\: R = \frac {u ^{2} sin2 \theta }{g }$

R = u² * 2 ( 4/5) ( 3/5) ÷ g

3 * 10^3 * 10 = 24/25 ( u²)

3 * 10^4 * 25 / 24 = u²

250000/8 = u²

125000/4 = u²

u² = 31250

u² = 15625 * 2

u² = (125)² * (√2)² m/s

u = 125√2 m/s

Therefore, The Velocity of projection is 125√2 m/s.