Quote (2wo1ne @ Feb 10 2015 11:04pm)
Well for the 1st question u can break the initial velocity into its vertical and horizontal component
horizontal velocity, v_x = v_i * cos (theta)
vertical velocity, v_y = v_i * sin(theta)
neglecting drag, v_x is constant and you know distance so you can find time
d/(v_x) = t
d/ v_i * cos (theta) = t
now use the equation h= v_y * t + (1/2)* at^2 for the vertical component
u know t = d/v_i * cos (theta), a = -g, v_y = v_i * sin(theta)
plug into equation and now you get
h = v_i * sin(theta) * d/v_i * cos (theta)+ (1/2)* (-g) (d/v_i * cos (theta))^2
u can probably simplify but im 2 lazy to try
this works, or you can just find the time from t = d/v_i * cos (theta)
then use that time to solve in your height equation h= v_y * t + (1/2)* at^2
If h is positive (congrats, you hit the building, if not, you fell short.