Question

A particle is projected with a velocity of u so that its horizontal range is twice the greatest height attainex the horizontal range is

Answer 1

Horizontal Range=u2 sin2θ/g

Max height =u2 sin^2 θ/2g

Now , 2 Max ht= Range

2×u2 sin^2θ/2g=u2sin2θ/g

Sin^2 θ=sin 2θ