Answer:
The launch angle [tex]\theta=45^o[/tex].
Explanation:
We know,
Horizontal distance travelled in projectile motion , [tex]R=\dfrac{u^2sin2\theta}{g}[/tex].
Also, Maximum height ,[tex]H_{max}=\dfrac{u^2sin^2\theta}{2g}[/tex].
Now, according to question maximum height is 1/4 of its horizontal range.
[tex]H_{max}=\dfrac{1}{4}\times R[/tex]
[tex]\dfrac{u^2 sin^2\theta}{2g}=\dfrac{1}{4}\times \dfrac{u^2 sin\ 2\theta}{g}[/tex]
Simplifying above expression.
We get , [tex]\theta=45^o[/tex]
Hence, this is the required solution.
