Question

10. A ball is projected with the velocity V so that its range on a horizontal plane is twice the greatest height,
then its range will be:​

Answer 1

Answer:

range= v^2sin 2theta/g

max height= v^2/2g

given , v^2sin2 theta/g=2(v^2/2g)

           sin 2 theta=1

           theta= pi/4

so range=v^2 sin 2(pi/4)/g

            =v^2/g

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