Question

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

Answer 1

Here's answer of the question

Answer 2

As we know horizontal range =(ucostheta)t

And t =2usintheta/g

where u is initial velocity

so horizontal range=(ucostheta)×t

=2u^2×sin2theta/g

At maximum height v=0

v^2=u^2+2as(3rd equation) where u^2 =usintheta(we will use only the vertical component of projectile motion)

so maximum height s =(usintheta)^2/2g

Now coming to the question,

horizontal range = 4maximum height

=2u^2sin2theta/g=4(usintheta)^2/2g

solving we will get

sin2theta=2sin^2theta

=2sintheta×costheta=2sin^2theta

sintheta/costheta=tantheta=1

=theta=1