Question

The velocity of projection is 6i+8j m/s the horizontal range of projectile is

Answer 1

Explanation:

Explanation:u = (6 i + 8 j)

Explanation:u = (6 i + 8 j)|u| = (6^2 + 8^2)^1/2 = 10 m/s

Explanation:u = (6 i + 8 j)|u| = (6^2 + 8^2)^1/2 = 10 m/sSo,

Explanation:u = (6 i + 8 j)|u| = (6^2 + 8^2)^1/2 = 10 m/sSo,ux = 6 m/s

Explanation:u = (6 i + 8 j)|u| = (6^2 + 8^2)^1/2 = 10 m/sSo,ux = 6 m/suy = 8 m/s

Explanation:u = (6 i + 8 j)|u| = (6^2 + 8^2)^1/2 = 10 m/sSo,ux = 6 m/suy = 8 m/sAngle of projection is θ .

Explanation:u = (6 i + 8 j)|u| = (6^2 + 8^2)^1/2 = 10 m/sSo,ux = 6 m/suy = 8 m/sAngle of projection is θ .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

Answer 2

I hope you can understand this