
A bullet fired from
A with a velocity of 900 ft/s must hit the target at B. Given the position
of B with respect to A, find the angle q with
which the bullet must be fired?
a. 30.5^{0},
65.5^{0}
b. 28.5^{0}, 59.5^{0}
c. 32.5^{0}, 68.5^{0}
d. 34.5^{0},
70.5^{0}


Let t be the time
taken for the bullet to reach B
The horizontal distance
travelled by the bullet is given by
x = v_{x} t
where x = 16000 ft and v_{x} = 900 cos q
t = x/v_{x}
The vertical distance
travelled is given by
y = v_{y(t)}  (1/2)(g)t^{2}
where y = 300 ft, v_{y} = 900 sin q
and g = 32.2 ft/s^{2}
=> 3000 = 16000 tan q  5088 (1 + tan^{2}
q)
tan^{2} q  3.145 tan q
+ 1.6 = 0
The roots of the above equation are
tan q = 0.638 and tan q
= 2.538
q = 32.5^{0} or 68.5^{0}
The bullet must be fired at an angle of 32.5^{0} or 68.5^{0}
