Question

What is the value of tan theta if horizontal range is equal to tge maximum height?

Answer 1

Answer:

Let angle of projection is ∅ , u is the intial velocity .

we know,

Maximum Height = u²sin²∅/2g

horizontal range = u²sin2∅/g

A/C to question ,

u²sin²∅/2g = u²sin2∅/g

[ use, sin2x = 2sinx.cosx ]

sin²∅/2 = 2sin∅.cos∅

tan∅ = 4

∅ = tan-¹( 4)

angle of projection = tan-¹(4)