Q1: If the horizontal range is 2.5 times the greatest height, the angle of projection of the projectile, is

ANS:B - 58°

Horizontal range,R= (u^2sin(2$))/g........(eq1)
Maximum height,H=(u^2sin^2($))/2g........(eq2)

Here,
Horizontal range= 2.5*Maximum height
R=2.5*H.....(eq3)

Substitute eq 1 & eq 2 in eq3

Sin(2$)=(2.5sin^2($))/2
2sin($)cos($)= (2.5sin^2($))/2
tan($)=1.6
$=57.99°=58°