Question

Show that the maximum height and range of a projectile launched at an angle of 450 is one quarter of its range.

Answer 1

the relationship is 1/4 range

Answer 2

We know that ,

Height=u²sin²theta/2g

Now ,theta=45°

So,H=u²sin²(45°)/2g

H=u²/4g. (As,sin²(45°)=1/2).

4H=u²/g. -eq.(1)

Now,range=u²sin2theta/g

R=u²sin(2*45°)/g

R=u²sin(90°)/g

R=u²/g. -eq.(2)

Comparing eq.(1) and eq.(2),we get that,

4H=R

So,H=1/4*R