posted by Brad on .
An arrow is shot from the top of a building (height of 30 meters). The arrow travels 100 meters in the horizontal direction from the base of the building to the point where it hits the ground in 4 seconds. Find the angle of release, actual velocity at release, and maximum height of the arrow.
I found horizontal velocity = 25m/s, but I don't know what to do next.
h = Hi +Vi t -4.9 t^2
0 = 30 + Vi * 4 -4.9 * 16
solve for Vi, the initial velocity component up
then tan angle = Vi/25