An archer shoots an arrow into the air with an initial velocity of 75 feet per second. The initial height of the shot is 5 feet. The arrow strikes the ground precisely 200 feet from the archer. At what angle did the archer shoot the arrow?
Choose the positive direction of vertical velocity "up", and the starting point of the arrow to be the origin.
Let& t& be the time of arrow fly, and& A be angle at which the archer shoot the arrow. Then the initial horisontal velocity of the arrow is & Vh = 75 cos(A), and the vertical one is & Vv = 75 sin(A).
Notice that the arrow moved horizontally with constant velocity& Vh& the time t and pass the distance 200 feet. Then
& 200 = t * 75 cos(A) whence & t = 200/(75 cos(A)) = 8/(3 cos(A)).
On the other hand, the arrow moves vertically with constant acceleration & g = -9.8 m/s = - 9.8 * 3.2808 ft/s = -32.15 ft/s and the initial velocity Vh. When it strikes the ground, its y-coordinate is -5, therefore we have the identity:
The left hand side is equal to sin(B+2A), so (*) sin(B+2A) =& 1.119541 which is impossible.
This means that the assumptions of the problem are NOT CONSISTENT and the described situation is NOT POSSIBLE.
On the other hand, if we increase the initial velocity (75) and the height (5) , and decrease the distance (200) then the problem would have the solution, and reduces to solution of (*) with respect to A