Question

If b is the horizontal range for threta inclination and h is the max height reached by the projectile show that then max range is given by r^2/8h + 2h

Answer 1

Answer:

Explanation:

Range(R)=u2sin(2a)/g

and 

Height(H)=u2sin2(a)/2g

where, u is the initial velocity or velocity of projection, a is the angle of projection and g is the acceleration due to gravity.

Using the identify, sin(2a)=2sin(a)cos(a) in the range formula and dividing by the height formula, we get the relation― tan(a)=4H/R …(1)

We also know an identify that― sin(2a)=2tan(a)/1+tan2(a)

Substitution for tan(a) from (1), willl give,

sin(2a)=8RH/R2+16H2

Putting this result back in the formula for range will give―

R=(u2)∗8RH/(R2+16H2)∗g

Answer  = u = √(R2+16H2)∗g/8H 

Answer 2

Explanation:

b=vo²sin²theta/g

h=vo²sin²theta /2g

and r max= vo²/g

Here,

h/R=sintheta/4cos theta

Little track=16h²/R²

1)16h²/R²+16h²=4h/√R²+16h²

2)R²+16h²/R²=R/√R²+16h²

And R max= 4/2.4h.R

Lastly R max= R²/8h+2h

Hope it will help you with