Jul 9, 2024
u_x = u cos(θ)
: Initial speed in x-direction.u_y = u sin(θ)
: Initial speed in y-direction.a_x = 0
: No acceleration in the x-direction.a_y = -g
: Gravity in the y-direction, always acting downwards.T = 2u sin(θ) / g
u_y = u sin(θ)
, Final Velocity v_y = 0
v = u + at
0 = u sin(θ) - gt
t = u sin(θ) / g
T = 2t
H = (u^2 sin^2(θ)) / (2g)
u_y = u sin(θ)
, Final Velocity v_y = 0
v² = u² + 2as
to solve for displacement H.R = (u^2 sin(2θ)) / g
R = u^2 (2 sin(θ) cos(θ)) / g
u_x = u cos(θ)
, No X-direction Acceleration.T = 2u sin(θ) / g
s = ut + 0.5at²
simplify to R = u² sin(2θ) / g
u_x = 4i
, u_y = 3j
H
and Range R
are equal.y = x tan(θ) - (g x²) / (2u² cos²(θ))