Question

found to Ground Projectile
The velocity of projection of a projectile is (6i+8j) ms-1. The horizontal range of the projectlei
(A) 4.9 m
(B) 9.6 m
(C) 19.6 m
(D) 14 m​

Answer 1

Answer:

The velocity is,

u = (6 i + 8 j)

|u| = (6^2 + 8^2)^1/2 = 10 m/s

So,

ux = 6 m/s

uy = 8 m/s

Angle of projection is θ (say).

tan θ = uy/ux = 8/6

Now, range R = u^2sin( 2θ)/g =2u^2sinθcosθ/g= 2×100×6/10×8/10×1/10=9.6m