Question

Derive an expression for maximum horizontal range when a projectile is projected with an angle theta and velocity be u

Answer 1

$\huge\mathfrak\pink{anSwEr}$

Maximum horizontal range :-

sin 2theta = 1

sin 2theta = sin 90°

theta = 45°

R = u² sin2 (45°)/g

R = u² sin 90° /g

R = u²/g

$\huge{\bf{\blue{\fbox{\underline{\color{orange}</p><p>{hoPe\:iT\:heLpS\:!! }}}}}}$

Answer 2

$\huge\red{hElLo}$

horizontal range = u² sin2Φ /g

where Φ is the angle with which the projectile is projected.

we know,

for maximum horizontal range, sin 2Φ = 1
sin2Φ = sin 90°
2Φ = 90°
Φ = 45° ........ put this value in formula.

R = u² sin2(45°)/g
R = u² sin 90° /g
R = u² /g

hope it helps u ❤