Question

I did the problem but i'm confused on what to do next:
1.A hunter aims directly at a target (on the same level) 120m away. (a) If the bullet leaves the gun at a speed of 250m/s, by how much will it miss the target? (b) At what angle should the gun be aimed so the target will be hit?

Solution. (a) We choose a coordinates system with the origin at the release point, with x horizontal and y vertical with the positive direction down. We find the time of flight from the horizontal motion.

x = x0 + v0t;

120m = 0 + (250m/s)t, which gives t = 0.480s.

We find the distance the bullet falls from

y = y0+ v0yt + 1/2at2;

y = 0 + 0 + 1/2(9.80m/s2)(0.480s)2 = 1.13m.

I don't get the delta, cosine, and sine

Answer 1

the tangent of the elevation angle is the bullet drop (1.13 m) divided by the target distance (120 m)

Answer 2

It fall more than a meter? checking

Yes, 1.129 m
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sorry, I am used to y up

Vi = 250 sin T
u = 250 cos T

t = 120/u = 120/250 cos T

h = Hi + Vi t - 4.9 t^2
here we on the same level so h = Hi
0 = Vi t -4.9 t^2
t (4.9 t - Vi) = 0
obviously it is at starting height at t = 0
so what we want is
Vi = 4.9 t
but we know
t = 120/250 cos T
and
Vi = 250 sin T
so
250 sin T = 4.9 *120/250 cos T

ok from there ?