Question

ONLY GENUINE ANSWERS PLZ

A projectile is thrown from a platform at a height 10 m above the ground with velocity of
20 m/sec. At what angle should the projectile be thrown to reach the farthest point from O, which is
vertically below the point from which it is thrown. [g-10 m/s]​

Answer 1

HELLO DEAR,

let height of platform be h

and velocity of projection is v

GIVEN:-

h = 10m

v = 20m/s

g = 10m/s²

so, the maximum value of range R = v√(v² + 2gh)/g

now, the angle of projection ∅ is,

tan∅ = v²/gR

=> tan∅ = v²/v√(v² + 2gh)

=> tan∅ = v/√(v² + 2gh)

=> tan∅ = 20/√(400 + 2×10×10)

=> tan∅ = 20/√(400 + 200)

=> tan∅ = 20/10√6

=> tan∅ = 2/√6 = √(2/3)

=> ∅ = tan-¹{√(2/3)}

I HOPE IT'S HELP YOU DEAR,

THANKS