Question

The maximum range of projectile is 2/root 3times actual range . What is the angle of projection for the actual range?

Answer 1

Maximum range = u^2/g
Actual Range = u^2 Sin(2x)/g

u^2/g = (2/sqrt(3)) * u^2 Sin(2x)/g
sqrt(3)/2 = Sin(2x)
Sin60 = Sin(2x)
x = 60/2
x = 30°

Angle of projection of actual range is 30°

Answer 2

H e y a !

Your answer -: 30°

SOLUTION-:

Maximum range = u^2g

Actual Range=u^2Sin(2x)/g

So,

u^2/g = (2/sqrt(3)) * u^2 Sin(2x)/g

sqrt(3)/2 = Sin(2x)

Sin60 = Sin(2x)

x = 60/2

x = 30°

=> Angle of projection of actual range is 30°.....

#hope it helps !