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
Answers
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
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