[Chanel911]

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

[Perito]

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.

[Chanel911]

Thanks a lot, Perito! :D