Answer :
so to get the maximum range the angle must be equal to 45°
using the formula for taking out height reached by a projectile
we know
[tex]H=\frac{u^{2}sin^{2}{A}}{2g} [/tex]
so
[tex]R=\frac{u^{2}sin{2A}}{g} [/tex]
[tex]R=\frac{2u^{2}sin{A}cos{A}}{g} [/tex]
[tex]R=\frac{4u^{2}sin{A}cos{A}}{2g} [/tex]
Putting H
We have
[tex]\frac{R}{4}=\frac{u^{2}sin^{2}{A}cos{A}}{2g*sin{A}} [/tex]
[tex]\frac{R}{4}=H*\frac{cos{A}}{sin{A}} [/tex]
Putting A = 45°
we get
[tex]H=\frac{R}{4}[/tex]
using the formula for taking out height reached by a projectile
we know
[tex]H=\frac{u^{2}sin^{2}{A}}{2g} [/tex]
so
[tex]R=\frac{u^{2}sin{2A}}{g} [/tex]
[tex]R=\frac{2u^{2}sin{A}cos{A}}{g} [/tex]
[tex]R=\frac{4u^{2}sin{A}cos{A}}{2g} [/tex]
Putting H
We have
[tex]\frac{R}{4}=\frac{u^{2}sin^{2}{A}cos{A}}{2g*sin{A}} [/tex]
[tex]\frac{R}{4}=H*\frac{cos{A}}{sin{A}} [/tex]
Putting A = 45°
we get
[tex]H=\frac{R}{4}[/tex]
Time to reach the maximum height H = T = u Sin θ / g
Range = R = (u Cos θ) * (2 * u sin θ)/g = u² Sin 2θ / g
Range is maximum when 2θ = 90⁰ and Sin 2θ = 1
R_max = R = u² / g
Maximum Height reached = H = u² Sin² θ / 2 g = u² / 4 g = R / 4
===================================
Formula for relation between maximum height and range is :
H / R = tan θ / 4
So H = R / 4 as tan 45⁰ = 1
Range = R = (u Cos θ) * (2 * u sin θ)/g = u² Sin 2θ / g
Range is maximum when 2θ = 90⁰ and Sin 2θ = 1
R_max = R = u² / g
Maximum Height reached = H = u² Sin² θ / 2 g = u² / 4 g = R / 4
===================================
Formula for relation between maximum height and range is :
H / R = tan θ / 4
So H = R / 4 as tan 45⁰ = 1