Question

Prove that a gun will shoot three times as high, when it
angle of elevation is 60" as when it is 30%, but cover the sam
horizontal range,​

Answer 1

Answer:

For the projectile motion, the maximum height attained is given by:

h=u2sin2θ2g

where, u = initial speed of bullet

g = acceleration due to gravity

θ = angle of projection

The height when the angle of elevation is 60o:

h1=u2sin260o2g=(3√2)2u22g=34u22g

The height when the angle of elevation is 30o:

h2=u2sin230o2g=(12)2u22g=14u22g

Therefore, the ratio of the two heights is given as:

h1h2=34u22g14u22g=31⇒h1=3h2

Now let us take the relation between the horizontal distance covered in the two cases:

Horizontal distance covered for a projectile is given as:

R=u2sin2θg

The horizontal distance covered when the angle of elevation is 60o:

R1=u2sin2×60og=u2sin120og=u2sin(90o+30o)g=u2cos30og=3√2u2g

The horizontal distance covered when angle of elevation is 30o:

R2=u2sin2×30og=u2sin60og=3√2u2g

Therefore, the ratio of the horizontal distance covered in the two cases:

R1R2=3√2u2g3√2u2g=1

therefore,R1=R2