A rifle that shoots bullets at 460m/s is to be aimed at a target 45.7m away at the level of the rifle. At what angle above the target must the rifle be pointed in order for the bullet to hit the target?

I got this far:
s(x) = V[i](x)t + 1/2at^2
= 460cos(theta)t + 1/2(0)t^2
= 460cos(theta)t... [equ1]

s(y) = V[i](y)t + 1/2at^2
0 = 460sin(theta)t + 1/2(-9.8)t^2
-4.9t^2 + 460sin(theta)t = 0... [equ2]

since V[i](x) = s(x)/t
t = s(x)/V[i](x)
= 45.7/460cos(theta)

-4.9t^2 + 460sin(theta)t = 0... [equ2]
-4.9(45.7/460cos(theta))^2 + 460sin(theta)(45.7/460cos(theta)) = 0
-0.048/cos^2(theta) + 45.7tan(theta) = 0
0.048/cos(theta) = 45.7sin(theta)


(0.048)(2) = (45.7) (sin(theta)cos(theta))(2)
sin(2)(theta) = (2(0.048))/45.7
(2)(theta) = sin^-1((2(0.048))/45.7)
(2)(theta) = 0.12 degrees
(theta) = 0.06 degrees


The answer can't be 0.06 degrees, could it? :confused:
Help Appreciated

Your method of solving the equation is correct and your answer is correct to the second decimal place.

45.7 meters isn't very far away. That's why you can shoot a rifle pretty straight.

Thanks a lot, Perito! :D