A projectile is fired with initial velicity V at an angle alpha to the horizontal. If the range is double the greatest height,
a) find the value of alpha
b) show that the range is 4V^2 divided by 5g
Show that the max range is when it is fired so that the range is 4 times the max height.
Derive:
y'' = - g
y' = - gt + Vsin α
y = - gt²/2 + Vtsin α
x'' = 0
x' = Vcos α
x = Vtcos α
y' = 0 occurs at the maximum height
=> t = Vsin α / g
=> y
max = - V²sin²α/2g + V²sin²α/g = V²sin²α/2g
Time of flight = 2Vsin α / g
=> Range = V²sin 2α / g
But range is twice the maximum height hence
V²sin 2α / g = V²sin²α/g
sin 2α = sin²α
sin α (2 cos α - sin α) = 0
α = tan
-12 for non-zero/non-boundary angles
Since tan α = 2, use a right-angled triangle to deduce
cos α = 1/√5
sin α = 2/√5
Hence sin 2α = 2 sin α cos α = 4/5
=> Range = V²(4/5) / g = 4V² / 5g
Now for the new problem we want the maximum range i.e. the maximum value of V²sin 2α / g which is clearly V² / g as the maximum of sin 2α is 1 which happens when α = π/4.
Max height = V²sin²α/2g
= V²/4g when α = π/4
Max range / Max height = 4
=> Max range is four times the max height under these conditions
