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
the tangent of the elevation angle is the bullet drop (1.13 m) divided by the target distance (120 m)
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 ?
To find the answer to part (b) of the problem, we need to understand the concepts of delta, cosine, and sine.
1. Delta (∆): In physics, the symbol ∆ (delta) represents change or difference. In this context, it refers to the change in height or distance.
2. Cosine (cos): The cosine of an angle in a right triangle is defined as the ratio of the adjacent side to the hypotenuse. In a coordinate plane, the cosine of an angle can be calculated using trigonometric functions. It helps us find the horizontal component of a vector.
3. Sine (sin): The sine of an angle in a right triangle is defined as the ratio of the opposite side to the hypotenuse. Similar to cosine, it helps us find the vertical component of a vector.
Now, let's apply these concepts to find the angle at which the gun should be aimed to hit the target:
We have already calculated that the bullet would miss the target by 1.13m vertically. To hit the target, we need to find the angle θ between the horizontal direction and the line connecting the release point and the target.
Let's break down the motion of the bullet into horizontal and vertical components.
Horizontal component:
The time of flight, which we found earlier, is t = 0.480s. Therefore, the horizontal distance covered by the bullet would be:
∆x = v0 * t = 250m/s * 0.480s = 120m
Vertical component:
The bullet falls by ∆y = 1.13m. Since the acceleration due to gravity acts in the opposite direction (downwards), we take the negative sign. So, ∆y = -1.13m.
Now, we can calculate the angle θ using the following equations:
∆x = ∆s * cos(θ)
120m = |∆s| * cos(θ) (Since we want the magnitude of ∆s, we use the absolute value.)
∆y = ∆s * sin(θ)
-1.13m = |∆s| * sin(θ)
Dividing both equations, we get:
tan(θ) = -1.13m / 120m
Using inverse tangent (tan^(-1)), we can find θ:
θ = tan^(-1)(-1.13m / 120m)
Plug in the values for ∆y and ∆x to calculate the angle θ using a scientific calculator or online calculator that has the tangent function. The resulting angle will tell you how much the gun needs to be aimed upwards from the horizontal direction to hit the target.