Question

at what suitable angle is the maximum height of the projectile 1/3 is equal to its range
plz give me proper answer with description ​

Answer 1

The range b = (vo^2*sin2a)/g, where vo is the initial speed and a is the launch angle.

The maximum height is hmax = (vo*sina)^2/2g

Hence (vo^2*sin2a)/3g = vo^2*(sina)^2/2g

That is (sin2a)/3 = (sina)^2/2

But sin2a = 2*sina*cosa

So (2*sina*cosa)/3 = (sina)^2/2

Hence tan a = 4/3.

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