1.

The velocity at the maximum height of a projectileis half of its initial velocity u. Its range on thehorizontal plane is22u(B) 2g3g22(D) 2g3g

Answer»

At maximum heightvelocity = uCosθu/2 = uCosθθ = 60°

R = u² Sin(2θ) / g= u² Sin(2 × 60°) / g= u²√3 / (2g)

Range on horizontal plane is u²√3 / (2g)



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