Projectile Motion
December 8, 2016
Categorised in: 1st Year Mechanics
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 $R_{max} = {u^2}/{g(1 + sinθ)}$
Pratik Kataria is currently learning Springboot and Hibernate.
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