Question

8. Show that range of an oblique projectile is maximum at angle 45° obtain relation between R max and H max​

Answer 1

Let Velocity is V and the angle of projection with horizontal is θ

so, Range(R)= (2V²sinθcosθ)/g

which can be written as R=(V²sin2θ)/g

now,

dR/dθ=0

⇒(V²/g)×2cos2θ=0

⇒2cos2θ=0

⇒cos2θ=0

⇒2θ=90

∴θ=45°.

now,

d²R/dθ²=-(V²/g)×4sin2θ

now putting the value of θ=45° in the above equation

we get, d²R/dθ²⇒ - (4V²/g) <0

Hence, at θ=45° ,Range is maximum.

R=u²sin2θ/g  AND H= u²sin²θ/2g

now , R/H=2sin2θ/sin²θ

               =2×2sinθcosθ/sin²θ

               =4cotθ

               =4cot45°

               =4  

R/H=4

R=4H  or H=R/4 is the relation between Rmax  and Hmax