An archer lying on the ground shoots an arrow at a target that is 400 meters away, and 40 meters high. The initial velocity of the arrow is 120 m/s At what angle does he need to point his bow in order to hit the target? (Assume the value of g as -9.81 m/s² ).

Answer 1

A = angle up

u = 120 cos A
400 = u t = 120 cos A * t
t = 400 / (120 cos A)

Vi = 120 sin A
v = Vi - 9.81 t

40 = Vi t -4.9 t^2

Answer 2

Still confused. What am I solving for? Is Vi the initial velocity? And wouldn't it have two components: horizontal and vertical?