Question

a particle is projected with velocity 6i + 8j ,2m away form vertical wall. after striking wall it land At
_____ away form wall b​

Answer 1

Answer:

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

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Explanation: