Question

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

Answer 1

Answer:

V = 125√2  m/s

Explanation:

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

Let say Initial Velocity = V  at an angle = Sin⁻¹(4/5)

Vertical Velcoity = VSin(Sin⁻¹(4/5)) = 4V/5

Time To reach Max height = 4V/5g  = 4V/50   (using g = 10)

Time of Flight = 2 * 4V/50   = 8V/50

Horizontal Velocity = VCos(Sin⁻¹(4/5)) = V √(1 - Sin²(Sin⁻¹(4/5))

= V √(1 - (4/5)²  =  3V/5

Horizontal Distance = 3V/5 * 8V/50

=> 24V²/250  = 3000

=> V = 125√2  m/s