Projectile Motion

For projectile on H.P.

⋆ Time of flight $t = {2u sinα}/{g}$

⋆ Horizontal Range $R = {u^2 sin2α}/{g}$

when α = 45°, Range is maximum: $R_{max} = {u^2}/{g}$



⋆ Max Height $H_{max} = {u^2 sin^2 α }/{2g}$

⋆ Equation of path (Trajectory) y = x tanα – ${g x^2}/{2 u^2 cos^2 α}$

⋆ When projectile is projected with a horizontal velocity from top of a tower of heigh h

$y = – {g x^2}/{2 u^2}$ [as α = 0]

-h = $-{g x^2}/{2 u^2}$

Horizontal distance, $x = v_o √{{2h}/{g}}$
Time of flight, t = $√{{2h}/{g}}$

⋆ Same range, same velocities, $α_1 + α_2 = 90°$

⋆ For a projectile velocity is minimum at top

$v_{min}$ = u cosα

⋆Over H.P. $R/H $= 4 cotα


⋆ Over HP, two projectiles projected with complementary angles α and 90 – α

Ratio: $H_1/H_2 = {sin^2 α}/{cos^2 α}$ and $t_1/t_2 = {sin α}/{cos α}$

⋆ Horizontal component of velocity u cosα = $R/t$

⋆ When K.E. of projectile is E then at highest point, K.E. = $(cosα)^2 E$
α = 45° -> $KE_{top} = E/2$

α = 30° -> $KE_{top} = {3E}/{4}$

α = 60° -> $KE_{top} = E/4$

⋆For a projectile KE is minimum at highest point and equals to

$KE_{min} = 1/2 m (v cosθ)^2$

⋆If initial velocity is increased by 10%, maximum range is increased by 21%

⋆ For a projectile, at any point $a_N = g cosθ$
θ is angle of velocity with horizontal at that point

⋆ At highest point, radius of curvature is minimum,

$ρ_{min} = {(u cosα)^2}/{g}$

Projectile on Inclined Plane:

θ = Angle made by plane with horizontal

α = Angle made by velocity with plane

Time of flight t = ${2u sinα}/{g cosθ}$

Range along the plane

R = ${2 u^2 sinα }/{g cos^2 θ} . cos(α + θ)$

Maximum Height, H = ${u^2 sin^2 α}/{2g cosθ}$



For Range to be maximum α = 45 – $θ/2$
and $R_{max} = {u^2}/{g(1 + sinθ)}$