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 is _______.the answer is 175 m/s.explain how?​

Answer 1

horizontal range , R = u²sin2θ/g

where θ is angle of elevation of projectile, u is velocity of projection and g is acceleration due to gravity.

here, R = 3km = 3000m , g = 9.8 m/s²

θ = sin^-1(4/5)

we know, sin2θ = 2sinθcosθ

= 2 × sin{sin^-1(4/5)} × cos{sin^-1(4/5)}

= 2 × 4/5 × cos{cos^-1(3/5)}

= 2 × 4/5 × 3/5

= 24/25

so, sin2θ = 24/25

then, 3000 = u²(24/25)/9.8

or, 3000 × 9.8 × 25/24 = u²

or, 1000 × 25/8 = u²

or, 125 × 9.8 × 25 = u²

or, u² = (25 × 25 × 49)

or, u = 175 m/s

hence, velocity of projection is 175m/s

Answer 2

Answer:

see the attached image

Explanation:

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