Question

find the angle of projection at which the horizontal range and maximum height of a projectile are equal.

Answer 1

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)

Answer 2

We know,

H(max) = u^2 sin^2 theta/2g

And

R= u^2 sin^2 theta/g

From question,

H = R

or, u^2 sin^2 theta/2g = u^2 sin^2 theta/h

or, sin^2 theta = 2(2 sin theta.cos theta)

or, sin^2 theta = 4 sin theta. cos theta

or, sin theta/ cos theta = 4

or, tan theta = 4

or, theta = tan^-1 (4)

i.e. Theta= 76 degrees (Answer)